answer：Alright, so I've got this problem about a circle and some points on it. Let me try to figure it out step by step. First, the circle is given by the equation ((x - a)^2 + y^2 = r^2), where (r > 0). There are four points: (P_1(1,1)), (P_2(0,2)), (P_3(1, sqrt{3})), and (P_4(1, -sqrt{3})). The problem says exactly three of these points lie on the circle (C). Part (1) asks for the equation of the circle (C). Hmm, okay. So, since exactly three points are on the circle, one of these four points isn't. I need to figure out which one it is and then find (a) and (r).Looking at the points, (P_3) and (P_4) are symmetric with respect to the x-axis because their y-coordinates are (sqrt{3}) and (-sqrt{3}). That makes me think that if one is on the circle, the other should be too, unless the circle isn't symmetric about the x-axis. But the general equation of the circle is ((x - a)^2 + y^2 = r^2), which is symmetric about the vertical line (x = a), not necessarily the x-axis. So, maybe both (P_3) and (P_4) are on the circle, but (P_1) or (P_2) is not.Wait, but the problem says exactly three points lie on the circle. So, either (P_1) or (P_2) is not on the circle. Let me check.If I assume that (P_3) and (P_4) are on the circle, then plugging them into the circle equation:For (P_3(1, sqrt{3})):[(1 - a)^2 + (sqrt{3})^2 = r^2]Which simplifies to:[(1 - a)^2 + 3 = r^2]For (P_4(1, -sqrt{3})):[(1 - a)^2 + (-sqrt{3})^2 = r^2]Which is the same as above, so that's consistent.Now, let's check (P_2(0,2)):[(0 - a)^2 + 2^2 = r^2]Which simplifies to:[a^2 + 4 = r^2]And (P_1(1,1)):[(1 - a)^2 + 1^2 = r^2]Which simplifies to:[(1 - a)^2 + 1 = r^2]So, if three points are on the circle, either (P_1) or (P_2) is not. Let's see if both (P_2) and (P_3) can be on the circle. If so, then (P_1) might not be.So, from (P_2): (a^2 + 4 = r^2)From (P_3): ((1 - a)^2 + 3 = r^2)Set them equal:[a^2 + 4 = (1 - a)^2 + 3]Expanding the right side:[a^2 + 4 = 1 - 2a + a^2 + 3]Simplify:[a^2 + 4 = a^2 - 2a + 4]Subtract (a^2 + 4) from both sides:[0 = -2a]So, (a = 0)Then, (r^2 = a^2 + 4 = 0 + 4 = 4), so (r = 2)So, the circle equation is (x^2 + y^2 = 4)Now, check if (P_1(1,1)) is on this circle:[1^2 + 1^2 = 1 + 1 = 2 neq 4]So, (P_1) is not on the circle, which fits the condition that exactly three points are on the circle. So, the equation is (x^2 + y^2 = 4).Okay, part (1) seems done.Part (2): Let line (l) with slope (k) pass through point (Q(4, -2)) but not through point (P_2). If line (l) intersects circle (C) at two distinct points (A) and (B).First, find the range of values for (k).So, the line passes through (Q(4, -2)) with slope (k), so its equation is:[y + 2 = k(x - 4)]Simplify:[y = kx - 4k - 2]Or:[kx - y - 4k - 2 = 0]We need this line to intersect the circle (x^2 + y^2 = 4) at two distinct points. So, the distance from the center of the circle to the line must be less than the radius.The center of the circle is at (0,0), radius is 2.The distance (d) from (0,0) to the line (kx - y - 4k - 2 = 0) is:[d = frac{|k(0) - 1(0) - 4k - 2|}{sqrt{k^2 + (-1)^2}} = frac{|-4k - 2|}{sqrt{k^2 + 1}}]This distance must be less than 2:[frac{|-4k - 2|}{sqrt{k^2 + 1}} < 2]Multiply both sides by (sqrt{k^2 + 1}):[|-4k - 2| < 2sqrt{k^2 + 1}]Square both sides to eliminate the absolute value and square root:[( -4k - 2 )^2 < 4(k^2 + 1)]Expand the left side:[16k^2 + 16k + 4 < 4k^2 + 4]Subtract (4k^2 + 4) from both sides:[12k^2 + 16k < 0]Factor:[4k(3k + 4) < 0]So, the inequality holds when (k) is between (-frac{4}{3}) and (0). But we also need to ensure that the line does not pass through (P_2(0, 2)). Let's check if (P_2) is on the line.Plug (x = 0) into the line equation:[y = k(0) - 4k - 2 = -4k - 2]Set this equal to 2:[-4k - 2 = 2 implies -4k = 4 implies k = -1]So, when (k = -1), the line passes through (P_2). Therefore, (k) cannot be (-1).Thus, the range of (k) is (-frac{4}{3} < k < 0), excluding (k = -1). So, in interval notation:[k in left( -frac{4}{3}, -1 right) cup left( -1, 0 right)]Okay, that's the range for (k).Now, the second part: Prove that the sum of the slopes of lines (P_2A) and (P_2B) is a constant value.So, (P_2) is at (0,2). Points (A) and (B) are the intersection points of line (l) and circle (C).Let me denote (A(x_1, y_1)) and (B(x_2, y_2)). The slopes of (P_2A) and (P_2B) are:[m_{P_2A} = frac{y_1 - 2}{x_1 - 0} = frac{y_1 - 2}{x_1}][m_{P_2B} = frac{y_2 - 2}{x_2}]We need to show that (m_{P_2A} + m_{P_2B}) is a constant.First, let's find the equations of line (l) and circle (C). Line (l) is (y = kx - 4k - 2). Substitute this into the circle equation:[x^2 + (kx - 4k - 2)^2 = 4]Expand the square:[x^2 + k^2x^2 - 8k^2x - 4k x + 16k^2 + 16k + 4 = 4]Wait, let me do that step by step.First, expand ((kx - 4k - 2)^2):[(kx - 4k - 2)^2 = (kx)^2 + (-4k - 2)^2 + 2(kx)(-4k - 2)][= k^2x^2 + (16k^2 + 16k + 4) + (-8k^2x - 4k x)]So, putting it back into the circle equation:[x^2 + k^2x^2 - 8k^2x - 4k x + 16k^2 + 16k + 4 = 4]Combine like terms:[(1 + k^2)x^2 + (-8k^2 - 4k)x + (16k^2 + 16k + 4 - 4) = 0]Simplify constants:[(1 + k^2)x^2 + (-8k^2 - 4k)x + 16k^2 + 16k = 0]Factor out 4 from the last two terms:[(1 + k^2)x^2 + (-8k^2 - 4k)x + 4(4k^2 + 4k) = 0]Hmm, not sure if that helps. Maybe just leave it as:[(1 + k^2)x^2 + (-8k^2 - 4k)x + 16k^2 + 16k = 0]Let me denote this quadratic in (x) as:[A x^2 + B x + C = 0]Where:[A = 1 + k^2][B = -8k^2 - 4k][C = 16k^2 + 16k]The solutions are (x_1) and (x_2), so by Vieta's formulas:[x_1 + x_2 = -frac{B}{A} = frac{8k^2 + 4k}{1 + k^2}][x_1 x_2 = frac{C}{A} = frac{16k^2 + 16k}{1 + k^2}]Now, since (y = kx - 4k - 2), we can express (y_1) and (y_2) in terms of (x_1) and (x_2):[y_1 = kx_1 - 4k - 2][y_2 = kx_2 - 4k - 2]So, the slopes are:[m_{P_2A} = frac{y_1 - 2}{x_1} = frac{kx_1 - 4k - 2 - 2}{x_1} = frac{kx_1 - 4k - 4}{x_1} = k - frac{4k + 4}{x_1}]Similarly,[m_{P_2B} = frac{y_2 - 2}{x_2} = frac{kx_2 - 4k - 4}{x_2} = k - frac{4k + 4}{x_2}]So, the sum:[m_{P_2A} + m_{P_2B} = 2k - (4k + 4)left( frac{1}{x_1} + frac{1}{x_2} right)]Now, let's compute (frac{1}{x_1} + frac{1}{x_2}):[frac{1}{x_1} + frac{1}{x_2} = frac{x_1 + x_2}{x_1 x_2}]We have expressions for (x_1 + x_2) and (x_1 x_2) from Vieta's formulas:[frac{x_1 + x_2}{x_1 x_2} = frac{frac{8k^2 + 4k}{1 + k^2}}{frac{16k^2 + 16k}{1 + k^2}} = frac{8k^2 + 4k}{16k^2 + 16k} = frac{4k(2k + 1)}{16k(k + 1)} = frac{4k(2k + 1)}{16k(k + 1)}]Simplify:[= frac{(2k + 1)}{4(k + 1)}]Assuming (k neq 0) and (k neq -1), which is true based on the range we found earlier.So, plugging back into the sum of slopes:[m_{P_2A} + m_{P_2B} = 2k - (4k + 4)left( frac{2k + 1}{4(k + 1)} right )]Simplify the second term:[(4k + 4) = 4(k + 1)]So,[(4k + 4)left( frac{2k + 1}{4(k + 1)} right ) = (k + 1)left( frac{2k + 1}{k + 1} right ) = 2k + 1]Therefore,[m_{P_2A} + m_{P_2B} = 2k - (2k + 1) = -1]So, the sum is (-1), which is a constant. That proves the second part.Wait, let me double-check the calculation:We had:[m_{P_2A} + m_{P_2B} = 2k - (4k + 4)left( frac{2k + 1}{4(k + 1)} right )]Simplify the fraction:[frac{4k + 4}{4(k + 1)} = frac{4(k + 1)}{4(k + 1)} = 1]Wait, that can't be right. Wait, no:Wait, actually, (4k + 4 = 4(k + 1)), so:[(4k + 4)left( frac{2k + 1}{4(k + 1)} right ) = 4(k + 1) cdot frac{2k + 1}{4(k + 1)} = 2k + 1]Yes, that's correct. So, the second term is (2k + 1), so:[2k - (2k + 1) = -1]Yes, that seems right. So, the sum is indeed (-1), a constant.So, I think that's the proof.**Final Answer**- (1) The equation of the circle (C) is (boxed{x^2 + y^2 = 4}).- (2) The range of values for (k) is (boxed{left( -frac{4}{3}, -1 right) cup left( -1, 0 right)}), and the sum of the slopes is the constant (boxed{-1}).

answer：Okay, so I have this triangle ABC where AB is 2 units, AC is 3 units, and the angle at A is 60 degrees. I need to find the length of BC and then the value of sin 2C. Hmm, let me think about how to approach this.First, for part (1), finding BC. I remember there's something called the Law of Cosines which relates the sides of a triangle to one of its angles. The formula is c² = a² + b² - 2ab cos(C), where C is the angle opposite side c. In this case, angle A is 60 degrees, and sides AB and AC are given. So, BC is the side opposite angle A. Let me write that down.So, BC² = AB² + AC² - 2 * AB * AC * cos(A). Plugging in the values, that would be BC² = 2² + 3² - 2 * 2 * 3 * cos(60°). Calculating each part: 2² is 4, 3² is 9, so 4 + 9 is 13. Then, 2 * 2 * 3 is 12, and cos(60°) is 0.5. So, 12 * 0.5 is 6. Therefore, BC² = 13 - 6, which is 7. So, BC is the square root of 7. That seems straightforward.Wait, let me double-check. AB is 2, AC is 3, angle A is 60 degrees. Using the Law of Cosines, yes, that should give BC. So, BC is sqrt(7). Got it.Now, moving on to part (2), finding sin 2C. Hmm, okay. I know that sin 2θ is equal to 2 sin θ cos θ, so maybe I can find sin C and cos C first and then multiply them accordingly. To find sin C, I might need to use the Law of Sines, which states that a / sin A = b / sin B = c / sin C. In this triangle, sides AB, BC, and AC correspond to angles C, A, and B respectively. Wait, actually, side AB is opposite angle C, side AC is opposite angle B, and side BC is opposite angle A.So, using the Law of Sines, AB / sin C = BC / sin A. We know AB is 2, BC is sqrt(7), and angle A is 60 degrees. So, plugging in, we get 2 / sin C = sqrt(7) / sin 60°. Solving for sin C, that would be sin C = (2 * sin 60°) / sqrt(7). Sin 60° is sqrt(3)/2, so sin C = (2 * sqrt(3)/2) / sqrt(7) = sqrt(3)/sqrt(7). To rationalize the denominator, that would be sqrt(21)/7.Okay, so sin C is sqrt(21)/7. Now, to find cos C, I can use the Pythagorean identity: sin² C + cos² C = 1. So, cos² C = 1 - sin² C. Let's calculate sin² C: (sqrt(21)/7)² is 21/49, which simplifies to 3/7. Therefore, cos² C = 1 - 3/7 = 4/7. Taking the square root, cos C is sqrt(4/7) = 2/sqrt(7). Again, rationalizing the denominator, that's 2 sqrt(7)/7.Now, since angle C is opposite side AB, which is shorter than side BC, angle C must be acute. So, cos C is positive, which makes sense. Therefore, sin 2C is 2 sin C cos C. Plugging in the values: 2 * (sqrt(21)/7) * (2 sqrt(7)/7). Let's compute that.First, multiply the constants: 2 * 2 is 4. Then, sqrt(21) * sqrt(7) is sqrt(147). Wait, sqrt(147) can be simplified. 147 is 49 * 3, so sqrt(49 * 3) is 7 sqrt(3). So, sqrt(21) * sqrt(7) is 7 sqrt(3). Therefore, the numerator is 4 * 7 sqrt(3), which is 28 sqrt(3). The denominator is 7 * 7, which is 49. So, sin 2C is 28 sqrt(3)/49. Simplifying that, 28 and 49 have a common factor of 7. 28 divided by 7 is 4, and 49 divided by 7 is 7. So, sin 2C is 4 sqrt(3)/7.Let me just recap to make sure I didn't make any mistakes. For part (1), using the Law of Cosines, I correctly applied the formula and got BC as sqrt(7). For part (2), using the Law of Sines, I found sin C as sqrt(21)/7, then found cos C as 2 sqrt(7)/7, and then used the double-angle identity to find sin 2C as 4 sqrt(3)/7. Everything seems to check out.I think that's all. I don't see any errors in my calculations, so I feel confident with these answers.**Final Answer**(1) The length of BC is boxed{sqrt{7}}.(2) The value of sin 2C is boxed{dfrac{4sqrt{3}}{7}}.

answer：Okay, so I have this problem here about a rational function, specifically 1 over q(x), and it says that this function has vertical asymptotes at x = -2 and x = 2. I remember that vertical asymptotes occur where the denominator is zero, provided the numerator isn't also zero there. Since the function is 1 over q(x), that means q(x) must be zero at x = -2 and x = 2. Alright, so q(x) is a quadratic polynomial. Quadratic means it's a degree 2 polynomial, so it should have two roots. Given the vertical asymptotes, those roots must be at x = -2 and x = 2. So, I can write q(x) in its factored form as q(x) = a(x + 2)(x - 2), where 'a' is some constant coefficient. Let me check that. If I plug in x = -2, then (x + 2) becomes zero, so q(-2) = 0, which is correct for a vertical asymptote. Similarly, plugging in x = 2 gives (x - 2) = 0, so q(2) = 0. That makes sense. Now, I need to find the value of 'a'. The problem also gives me that q(3) = 10. So, I can substitute x = 3 into my expression for q(x) and set it equal to 10 to solve for 'a'. Let me write that out: q(3) = a(3 + 2)(3 - 2) = a(5)(1) = 5a. And this is equal to 10. So, 5a = 10. Dividing both sides by 5, I get a = 2. So, substituting back into q(x), I have q(x) = 2(x + 2)(x - 2). Let me expand that to make sure it's a quadratic. Multiplying (x + 2)(x - 2) gives x^2 - 4, so q(x) = 2(x^2 - 4) = 2x^2 - 8. Let me verify that this works. Plugging in x = 3, q(3) = 2*(3)^2 - 8 = 2*9 - 8 = 18 - 8 = 10. That's correct. Also, plugging in x = -2 and x = 2, q(-2) = 2*(-2)^2 - 8 = 8 - 8 = 0, and q(2) = 2*(2)^2 - 8 = 8 - 8 = 0. So, the vertical asymptotes are indeed at x = -2 and x = 2. Wait, just to make sure I didn't make any mistakes in my calculations. Let me go through the steps again. 1. Vertical asymptotes at x = -2 and x = 2 imply q(x) has roots at these points, so q(x) = a(x + 2)(x - 2). 2. Expanding gives q(x) = a(x^2 - 4). 3. Given q(3) = 10, substitute x = 3: a(9 - 4) = 10 => 5a = 10 => a = 2. 4. Therefore, q(x) = 2x^2 - 8. Yes, that all seems correct. I don't think I made any errors in the algebra or substitution. I wonder if there's another way to approach this problem. Maybe by using the general form of a quadratic polynomial, which is q(x) = ax^2 + bx + c, and then setting up equations based on the given information. Let me try that approach to confirm my answer. Given that q(x) is a quadratic with roots at x = -2 and x = 2, that means it can be written as q(x) = a(x + 2)(x - 2) as before, which expands to q(x) = a(x^2 - 4). So, in standard form, that's q(x) = a x^2 - 4a. So, comparing to the general form q(x) = ax^2 + bx + c, we can see that b = 0 and c = -4a. Now, using the condition q(3) = 10: q(3) = a*(3)^2 + b*(3) + c = 9a + 3b + c = 10. But since b = 0 and c = -4a, substituting these in: 9a + 0 + (-4a) = 10 => 5a = 10 => a = 2. So, again, we get a = 2, and therefore c = -8. So, q(x) = 2x^2 - 8. This confirms my earlier result. I think I've covered all the bases here. I approached the problem by considering the roots given by the vertical asymptotes, wrote q(x) in factored form, expanded it, used the given condition to solve for the coefficient, and then verified my answer by plugging in the value and also by approaching the problem from the standard quadratic form. Just to make sure, let me also graph the function q(x) = 2x^2 - 8 mentally. It's a parabola opening upwards with vertex at (0, -8). It crosses the x-axis at x = -2 and x = 2, which aligns with the vertical asymptotes of 1/q(x). At x = 3, the value is 2*(9) - 8 = 18 - 8 = 10, which matches the given condition. Everything seems consistent. I don't think I missed anything here. The key was recognizing that the vertical asymptotes correspond to the roots of the denominator polynomial q(x), and then using the given value to find the leading coefficient. I guess another thing to consider is whether there could be any other factors or terms in q(x), but since it's specified as a quadratic polynomial, it can only have two roots, which we've accounted for. So, there's no room for additional factors or terms without changing the degree of the polynomial. In summary, the steps I took were:1. Identify the roots of q(x) based on the vertical asymptotes.2. Express q(x) in factored form with a leading coefficient.3. Expand the factored form to standard quadratic form.4. Use the given value q(3) = 10 to solve for the leading coefficient.5. Verify the solution by plugging values back into the polynomial and ensuring consistency with the given conditions.This systematic approach helped me arrive at the correct answer without any confusion or mistakes along the way. It's always good to check your work by approaching the problem from different angles, as I did here, to ensure that the solution is solid and without errors.

answer：First, I'll assign variables to represent the temperatures for each day: M for Monday, T for Tuesday, W for Wednesday, Th for Thursday, and F for Friday.From the problem, the average temperature for Monday, Tuesday, Wednesday, and Thursday is 48 degrees. This means the total temperature for these four days is 48 multiplied by 4, which equals 192 degrees.Similarly, the average temperature for Tuesday, Wednesday, Thursday, and Friday is 46 degrees. Therefore, the total temperature for these four days is 46 multiplied by 4, which equals 184 degrees.I know that the temperature on Monday (M) is 39 degrees. I'll use this information to find the sum of the temperatures for Tuesday, Wednesday, and Thursday (T + W + Th). By subtracting Monday's temperature from the total of the first four days, I get T + W + Th = 192 - 39 = 153 degrees.Now, I'll use the total temperature for Tuesday, Wednesday, Thursday, and Friday (184 degrees) and subtract the sum of Tuesday, Wednesday, and Thursday (153 degrees) to find the temperature on Friday (F). This calculation gives F = 184 - 153 = 31 degrees.Therefore, the temperature on Friday was 31 degrees.

answer：First, I need to determine the total number of ceilings Michelangelo has to paint in the 4-story building. Each floor has 7 rooms, so the total number of room ceilings is 4 floors multiplied by 7 rooms per floor, which equals 28 room ceilings.Next, I need to account for the hallway ceilings. There is one hallway per floor, and with 4 floors, this adds 4 hallway ceilings.Adding the room ceilings and hallway ceilings together, the total number of ceilings to paint is 28 room ceilings plus 4 hallway ceilings, totaling 32 ceilings.This week, Michelangelo painted 12 ceilings. Subtracting this from the total, there are 32 total ceilings minus 12 painted ceilings, leaving 20 ceilings remaining.For next week, Michelangelo plans to paint one-fourth of the number of ceilings he painted this week, which is one-fourth of 12, equaling 3 ceilings. Additionally, he will paint the 4 hallway ceilings.Therefore, next week, Michelangelo will paint a total of 3 ceilings plus 4 hallway ceilings, totaling 7 ceilings.Subtracting the 7 ceilings he plans to paint next week from the remaining 20 ceilings, there will be 13 ceilings left to paint after next week.

answer：Okay, so I have this geometry problem here about a sphere and a pyramid. Let me try to understand it step by step. First, the problem says there's a sphere O with a radius of 3√2. On the surface of this sphere, there are four points: S, A, B, and C. We're told that angle ABC is 90 degrees, and both AB and BC are 4√2 units long. We need to find the maximum volume of the pyramid S-ABC. The options are given as A, B, C, D with different expressions involving √2.Alright, so let me break this down. I know that the volume of a pyramid is (1/3) times the base area times the height. So, to find the maximum volume, I need to maximize either the base area, the height, or both. Given that ABC is a right-angled triangle at B, with AB = BC = 4√2, I can find the length of AC. Since ABC is a right-angled triangle, AC should be the hypotenuse. So, AC = √[(4√2)^2 + (4√2)^2]. Let me compute that:(4√2)^2 = 16 * 2 = 32. So, AC = √(32 + 32) = √64 = 8. So, AC is 8 units long.Now, points A, B, and C are all on the surface of the sphere with radius 3√2. So, the distance from the center of the sphere O to each of these points is 3√2.I need to figure out how to place point S on the sphere such that the volume of pyramid S-ABC is maximized. Since the volume depends on the height from S to the base ABC, I should maximize this height.But how do I find the maximum possible height? Hmm. Maybe I can think about the position of S relative to the base ABC.First, let's consider the base ABC. Since ABC is a right-angled triangle with sides 4√2, 4√2, and 8, it's an isosceles right-angled triangle. So, the base is a triangle in 3D space, lying on the sphere.I think the key here is to find the position of S such that the height from S to the plane ABC is as large as possible. Since all points are on the sphere, the maximum height would occur when S is as far as possible from the plane ABC.To find the maximum height, I might need to find the distance from the center of the sphere O to the plane ABC. Then, the maximum height from S to the plane ABC would be the sum of the radius and this distance, or the difference, depending on the direction.Wait, actually, the maximum distance from a point on the sphere to the plane ABC would be the distance from the center to the plane plus the radius of the sphere. Because if the center is at a certain distance from the plane, then the farthest point on the sphere from the plane would be in the direction away from the plane, adding the radius to that distance.So, let me compute the distance from the center O to the plane ABC.But how do I find the distance from the center to the plane ABC? I think I need to find the equation of the plane ABC and then compute the distance from O to this plane.Alternatively, since ABC is a triangle on the sphere, maybe there's a way to find the distance using the properties of the sphere and the triangle.Wait, another approach: since ABC is a triangle on the sphere, the plane ABC is a great circle if ABC is a right-angled triangle? Hmm, not necessarily. A great circle is the intersection of the sphere with a plane passing through the center. So, unless the plane ABC passes through the center, it's not a great circle.But in this case, ABC is a triangle with sides 4√2, 4√2, and 8. Let's see if the triangle ABC is a great circle. The length of AC is 8, which is the diameter of the sphere? Wait, the sphere has radius 3√2, so diameter is 6√2, which is approximately 8.485. But AC is 8, which is less than the diameter. So, AC is not a diameter, meaning that the plane ABC does not pass through the center of the sphere.Therefore, the distance from the center O to the plane ABC is not zero. So, we need to compute this distance.How can I compute the distance from the center to the plane ABC?One method is to use the formula for the distance from a point to a plane in 3D space. But to use that, I need the equation of the plane ABC. To get the equation, I need coordinates of points A, B, and C.Alternatively, maybe I can use some geometric properties.Let me think. Since ABC is a right-angled triangle, with AB = BC = 4√2, and AC = 8, as we found earlier. So, triangle ABC is a right-angled isosceles triangle.If I consider the sphere with radius 3√2, and points A, B, C on its surface, then OA = OB = OC = 3√2.So, triangle ABC is inscribed in the sphere. The circumradius of triangle ABC can be found, and then we can relate it to the distance from the center O to the plane ABC.Wait, the circumradius of triangle ABC is the radius of the circle in which ABC is inscribed, which is the intersection of the sphere with the plane ABC. So, the circumradius of triangle ABC is equal to the radius of the sphere times the sine of the angle between the center and the plane.Wait, maybe it's better to recall that the distance d from the center O to the plane ABC is related to the circumradius R of triangle ABC and the sphere's radius r by the formula:d = √(r² - R²)So, if I can find the circumradius R of triangle ABC, then I can compute d.Yes, that seems promising.So, first, let's find the circumradius R of triangle ABC.For a right-angled triangle, the circumradius is half the hypotenuse. Since ABC is a right-angled triangle at B, the hypotenuse is AC, which is 8. Therefore, R = AC / 2 = 8 / 2 = 4.So, the circumradius R is 4.Given that the sphere has radius r = 3√2, then the distance from the center O to the plane ABC is d = √(r² - R²) = √[(3√2)² - 4²] = √[18 - 16] = √2.So, d = √2.Therefore, the distance from the center O to the plane ABC is √2.Now, to find the maximum height of the pyramid S-ABC, we need to find the maximum distance from point S to the plane ABC.Since S is on the sphere, the maximum distance from S to the plane ABC will be the distance from the center O to the plane ABC plus the radius of the sphere in the direction away from the plane.Wait, actually, the maximum distance from S to the plane ABC is d + r, but wait, that would be if S is on the opposite side of the plane from O. But actually, the maximum distance is r + d if S is on the same side as O, but I think it's actually the other way around.Wait, no, let me think carefully.If the center O is at a distance d from the plane ABC, then the maximum distance from any point S on the sphere to the plane ABC would be d + r, but only if the point S is on the line through O perpendicular to the plane ABC, on the opposite side from the plane.Wait, actually, no. Because if O is at distance d from the plane, then the farthest point S on the sphere from the plane ABC would be along the line through O perpendicular to the plane ABC, on the side opposite to the plane. So, the distance from S to the plane ABC would be d + r.Wait, but wait, the sphere has radius r, so the distance from O to S is r. So, if O is at distance d from the plane, then S can be at most d + r away from the plane, but only if S is in the direction away from the plane.But actually, in 3D space, the maximum distance from S to the plane ABC is d + r, but only if S is on the line through O perpendicular to the plane ABC, on the side opposite to the plane.Wait, let me verify this.Imagine the plane ABC is at a distance d from O. Then, the sphere is centered at O with radius r. The maximum distance from any point S on the sphere to the plane ABC would be the distance from O to the plane plus the radius in the direction away from the plane. So, yes, it's d + r.But wait, in our case, d is √2, and r is 3√2. So, the maximum distance from S to the plane ABC would be √2 + 3√2 = 4√2.Wait, but hold on, is that correct? Because if the center is at distance d from the plane, then the maximum distance from S to the plane would be d + r, but only if S is on the side of O opposite to the plane. If S is on the same side as O, then the distance would be |d - r|, but since r > d, that would be r - d.But in our case, we want the maximum distance, so it's d + r.Wait, but actually, no. Because the distance from S to the plane can't exceed r + d, but in reality, the maximum occurs when S is diametrically opposite to the closest point on the plane.Wait, perhaps I should think in terms of vectors.Let me consider the plane ABC, and the center O at distance d from the plane. Then, the maximum distance from S to the plane ABC is the distance from O to the plane plus the maximum possible component of S in the direction away from the plane.Since S is on the sphere, the maximum occurs when S is along the line through O perpendicular to the plane ABC, on the side away from the plane. So, the distance from S to the plane would be d + r.Wait, but that would mean the distance is √2 + 3√2 = 4√2, as I thought earlier.But wait, let me think again. If O is at distance d from the plane, then the point on the sphere farthest from the plane would be in the direction away from the plane, so the distance would be d + r.But wait, actually, no. Because the distance from O to the plane is d, so the point on the sphere in the direction away from the plane would be at distance d + r from the plane. But wait, the sphere's radius is r, so the distance from O to S is r, so the maximum distance from S to the plane is d + r.Wait, but in our case, d is √2, and r is 3√2, so the maximum distance is √2 + 3√2 = 4√2.But wait, let me confirm this with an example. Suppose the sphere has radius 1, and the plane is at distance 0.5 from the center. Then, the maximum distance from a point on the sphere to the plane would be 0.5 + 1 = 1.5, which is correct.Similarly, if the plane is at distance d from the center, then the maximum distance from a point on the sphere to the plane is d + r.So, in our case, d = √2, r = 3√2, so maximum distance is 4√2.Therefore, the maximum height h of the pyramid S-ABC is 4√2.Now, the volume of the pyramid is (1/3) * base area * height.We need to compute the base area, which is the area of triangle ABC.Since ABC is a right-angled triangle with legs 4√2 each, the area is (1/2) * AB * BC = (1/2) * 4√2 * 4√2.Let me compute that:4√2 * 4√2 = 16 * 2 = 32.So, area = (1/2) * 32 = 16.Therefore, the base area is 16.So, the volume V = (1/3) * 16 * h, where h is 4√2.So, V = (1/3) * 16 * 4√2 = (64√2)/3.Looking at the options, option A is 64√2 / 3, which matches our result.Wait, but let me double-check my reasoning.First, I found that AC = 8, which is correct because in a right-angled triangle with legs 4√2, the hypotenuse is 8.Then, I found the circumradius R of triangle ABC, which for a right-angled triangle is half the hypotenuse, so R = 4. That seems correct.Then, using the formula d = √(r² - R²), where r is the sphere's radius, 3√2, so r² = 18, R² = 16. Therefore, d = √(18 - 16) = √2. That seems correct.Then, the maximum height h is d + r = √2 + 3√2 = 4√2. That seems correct.Then, the area of the base ABC is 16, as computed. So, volume is (1/3)*16*4√2 = 64√2 / 3, which is option A.But wait, let me think again about the maximum height. Is it really d + r?Wait, actually, if the center is at distance d from the plane, then the maximum distance from a point on the sphere to the plane is r + d, but only if the point is on the opposite side of the plane from the center.Wait, but in our case, the center is at distance d = √2 from the plane, so the point S on the sphere that is farthest from the plane would be in the direction away from the plane, so the distance from S to the plane would be d + r, but wait, that would be √2 + 3√2 = 4√2, which is correct.But wait, actually, the distance from S to the plane is the distance from O to the plane plus the component of OS in the direction perpendicular to the plane.Since OS is a radius, its length is r = 3√2. The maximum occurs when OS is in the same direction as the normal vector from O to the plane, so the distance from S to the plane is d + r.Wait, but actually, no. Because the distance from S to the plane is |distance from O to plane ± component of OS along the normal|.Wait, perhaps I should think in terms of vectors.Let me denote the normal vector to the plane ABC as n. The distance from O to the plane is d = √2. So, the vector from O to the plane is d in the direction of n.Then, the point S on the sphere can be written as O + r * (n / |n|), but normalized.Wait, actually, the maximum distance occurs when S is in the direction of n from O, so S = O + r * (n / |n|). But since n is a unit vector, S = O + r * n.Wait, but the distance from S to the plane is then |(O + r * n) · n - D|, where D is the scalar in the plane equation. But since the distance from O to the plane is d, then D = -d (assuming the plane equation is n · x + D = 0, and O is at distance d from the plane).Wait, perhaps this is getting too complicated.Alternatively, think of it as the distance from S to the plane is the distance from O to the plane plus the projection of OS onto the normal vector.Since S is on the sphere, OS has length r. The maximum projection occurs when OS is in the same direction as the normal vector, so the projection is r. Therefore, the maximum distance from S to the plane is d + r.Wait, but in our case, d is the distance from O to the plane, which is √2, and r is 3√2, so the maximum distance is √2 + 3√2 = 4√2.Therefore, the height h is 4√2.So, the volume is (1/3)*base area*height = (1/3)*16*4√2 = (64√2)/3, which is option A.But wait, let me check if there's another way to get a larger volume. Maybe if S is not on the line through O perpendicular to the plane ABC, but somewhere else. But I think the maximum height occurs when S is as far as possible from the plane, which is along the normal direction.Therefore, I think the maximum volume is indeed (64√2)/3, which is option A.But wait, let me think again. The base area is 16, and the height is 4√2, so volume is (1/3)*16*4√2 = 64√2 / 3.Yes, that seems correct.Alternatively, maybe I made a mistake in computing the base area. Let me check.Triangle ABC is right-angled at B, with AB = BC = 4√2. So, area is (1/2)*AB*BC = (1/2)*(4√2)*(4√2) = (1/2)*16*2 = (1/2)*32 = 16. That's correct.So, all steps seem correct. Therefore, the maximum volume is 64√2 / 3, which is option A.