answer:- **Calculate the Center**: The center of circle C is the midpoint of the segment from (2, -3) to (8, 9). Using the midpoint formula: [ left( frac{2+8}{2}, frac{-3+9}{2} right) = left( frac{10}{2}, frac{6}{2} right) = (5, 3) ] So, the center of circle C is (5, 3). - **Calculate the Radius**: The radius of the circle can be calculated using the distance formula between the center (5, 3) and one of the endpoints, for example, (2, -3): [ sqrt{(2-5)^2 + (-3-3)^2} = sqrt{(-3)^2 + (-6)^2} = sqrt{9 + 36} = sqrt{45} = 3sqrt{5} ] So, the length of the radius of circle C is 3sqrt{5}. Conclusion: The coordinates of the center of circle C are (5, 3) and the length of the radius is 3sqrt{5}. The final answer is: [ boxed{(5, 3), 3sqrt{5}} ]

answer:First, rewrite the equation of the circle in standard form: (x+ frac{a}{2})^{2}+(y+ frac{b}{2})^{2}= frac{a^{2}+b^{2}}{4}. Thus, the coordinates of the circle's center are (-frac{a}{2},-frac{b}{2}), and its radius is r= frac{sqrt{a^{2}+b^{2}}}{2}. Now, calculate the distance d between the center of the circle and the line ax+by=0: d= frac{frac{a^{2}+b^{2}}{2}}{sqrt{a^{2}+b^{2}}}= frac{sqrt{a^{2}+b^{2}}}{2} = r. Since the distance d between the center of the circle and the line is equal to the radius r of the circle, the line and the circle are tangent to each other. Therefore, the correct answer is: boxed{B}. To solve this problem, we first transformed the circle's equation into its standard form to find the center and radius. Then, we used the formula for the distance between a point and a line to calculate the distance between the center of the circle and the given line. By comparing the calculated distance with the circle's radius, we determined that the line is tangent to the circle. This problem involves understanding the relationship between a line and a circle, including the standard form of a circle's equation, the formula for the distance between a point and a line, and the property that when the distance between the center of a circle and a line is equal to the circle's radius, the line is tangent to the circle. Mastering this property is essential for solving this problem.

answer:1. **Define and substitute ( c ):** Let ( c = frac{2pi}{13} ). Substitute: [ frac{sin frac{8pi}{13} cdot sin frac{16pi}{13} cdot sin frac{24pi}{13} cdot sin frac{32pi}{13} cdot sin frac{40pi}{13}} {sin frac{2pi}{13} cdot sin frac{4pi}{13} cdot sin frac{6pi}{13} cdot sin frac{8pi}{13} cdot sin frac{10pi}{13}} ] 2. **Use properties of sine:** - (sin frac{16pi}{13} = sin (2pi - frac{16pi}{13}) = sin frac{10pi}{13}) - (sin frac{24pi}{13} = sin (2pi - frac{24pi}{13}) = sin frac{12pi}{13} = sin (frac{12pi}{13} - 2pi) = sin frac{-12pi}{13} = -sin frac{12pi}{13}) - (sin frac{32pi}{13} = sin (2pi - frac{32pi}{13}) = sin frac{6pi}{13}) - (sin frac{40pi}{13} = sin (2pi - frac{40pi}{13}) = sin frac{4pi}{13}) 3. **Substitute back and simplify:** [ frac{(-sin frac{8pi}{13} cdot sin frac{10pi}{13} cdot -sin frac{6pi}{13} cdot sin frac{4pi}{13})} {sin frac{2pi}{13} cdot sin frac{4pi}{13} cdot sin frac{6pi}{13} cdot sin frac{8pi}{13} cdot sin frac{10pi}{13}} ] Every term in the numerator has a corresponding matching term in the denominator canceling out except for ( sin frac{2pi}{13} ). 4. **Conclude with the final answer:** [ frac{1{sin frac{2pi}{13}}} ] The final answer is boxed{B) ( frac{1}{sin frac{2pi}{13}} )}

answer:We are given a point M inside the triangle ABC from which perpendiculars are dropped to the three sides, meeting at points on each side. We are to prove that the length of these perpendiculars (which are equal) is the diameter of the incircle of triangle ABC. 1. **Define the Altitudes and Perpendiculars:** Let AD, BE, and CF be the altitudes of triangle ABC, with M being the point inside the triangle such that the perpendiculars from M to these altitudes meet at I, J, and K respectively. Suppose AI = BJ = CK = t. 2. **Express the Area of Triangle ABC in Two Ways:** The area S_{triangle ABC} can be expressed using any of the altitudes: [ S_{triangle ABC} = frac{1}{2} AB cdot CF = frac{1}{2} BC cdot AD = frac{1}{2} CA cdot BE ] Additionally, S_{triangle ABC} can be expressed in terms of point M: [ S_{triangle ABC} = S_{triangle ABM} + S_{triangle BCM} + S_{triangle CAM} ] Hence, [ S_{triangle ABC} = frac{1}{2} AB cdot FK + frac{1}{2} BC cdot DI + frac{1}{2} CA cdot EJ ] 3. **Simplify Using Perpendiculars and Altitudes:** Since FK, DI, and EJ are respectively the perpendiculars from M to AD, BE, CF, and we are given AI = BJ = CK = t, we can rewrite the terms using t as follows: [ 2S_{triangle ABC} = AB cdot CF + BC cdot AD + CA cdot BE - (AB cdot FK + BC cdot DI + CA cdot EJ) ] Which simplifies to: [ AB cdot CK + BC cdot AI + CA cdot BJ = frac{t}{2} (AB + BC + CA) ] 4. **Relate Areas with the Inradius:** If r is the inradius of triangle ABC, then: [ S_{triangle ABC} = frac{1}{2} r (AB + BC + CA) ] 5. **Compare the Two Area Expressions:** Equating the two expressions obtained from steps 3 and 4: [ frac{t}{2} (AB + BC + CA) = frac{r}{2} (AB + BC + CA) ] Since AB, BC, and CA are non-zero, we can conclude: [ t = 2r ] Therefore, the length t of the perpendicular from point M to any of the altitudes (which are equal) is exactly twice the inradius r of triangle ABC. Thus, t equals the diameter of the incircle. [ boxed{t} ]