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(x+5)^(2)+(y-3)^(2)=169 In the xy-plane, the graph of the equation is a circle. Point A is on the circle and has coordinates (7,8). If bar(AB) is a diameter of the circle, what are the coordinates of point B ? Choose 1 answer: (A) (-17,-2) (B) (-5,3) (c) (-5,16) (D) (1,(11)/(2))

(x+5)2+(y−3)2=169 (x+5)^{2}+(y-3)^{2}=169 \newlineIn the xy x y -plane, the graph of the equation is a circle. Point A A is on the circle and has coordinates (7,8) (7,8) . If AB‾ \overline{A B} is a diameter of the circle, what are the coordinates of point B B ?\newlineChoose 11 answer:\newline(A) (−17,−2) (-17,-2) \newline(B) (−5,3) (-5,3) \newline(C) (−5,16) (-5,16) \newline(D) (1,112) \left(1, \frac{11}{2}\right)

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Q. (x+5)2+(y−3)2=169 (x+5)^{2}+(y-3)^{2}=169 \newlineIn the xy x y -plane, the graph of the equation is a circle. Point A A is on the circle and has coordinates (7,8) (7,8) . If AB‾ \overline{A B} is a diameter of the circle, what are the coordinates of point B B ?\newlineChoose 11 answer:\newline(A) (−17,−2) (-17,-2) \newline(B) (−5,3) (-5,3) \newline(C) (−5,16) (-5,16) \newline(D) (1,112) \left(1, \frac{11}{2}\right)
  1. Circle equation and center: The equation (x+5)2+(y−3)2=169(x+5)^{2}+(y-3)^{2}=169 represents a circle with a radius of 1313 (since 169169 is 1313 squared) and a center at (−5,3)(-5,3) (the opposite of the values inside the parentheses).
  2. Finding point B: Point A (7,8)(7,8) lies on the circle. To find point B, which is at the opposite end of the diameter, we need to use the fact that the diameter passes through the center of the circle. The midpoint of the diameter is the center of the circle.
  3. Midpoint formula: The coordinates of the center of the circle are (−5,3)(-5,3). Since AA and BB are endpoints of a diameter, the center of the circle is the midpoint between AA and BB. We can use the midpoint formula to find the coordinates of BB: Midpoint =(x1+x22,y1+y22)= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).
  4. Calculating x-coordinate of B: Let's denote the coordinates of point B as (xB,yB)(x_B, y_B). Using the midpoint formula and the coordinates of point A (7,8)(7,8) and the center (−5,3)(-5,3), we have:\newline−5+72=xB+72\frac{-5 + 7}{2} = \frac{x_B + 7}{2} and 3+82=yB+82.\frac{3 + 8}{2} = \frac{y_B + 8}{2}.
  5. Calculating y-coordinate of B: Solving the equations from the previous step, we get:\newline1=xB+721 = \frac{x_B + 7}{2} and 5.5=yB+825.5 = \frac{y_B + 8}{2}.
  6. Coordinates of point B: Multiplying both sides of the first equation by 22, we get:\newline2=xB+72 = x_B + 7. Subtracting 77 from both sides, we find xB=−5x_B = -5.
  7. Coordinates of point B: Multiplying both sides of the first equation by 22, we get:\newline2=xB+72 = x_B + 7. Subtracting 77 from both sides, we find xB=−5x_B = -5.Multiplying both sides of the second equation by 22, we get:\newline11=yB+811 = y_B + 8. Subtracting 88 from both sides, we find yB=3y_B = 3.
  8. Coordinates of point B: Multiplying both sides of the first equation by 22, we get:\newline2=xB+72 = x_B + 7. Subtracting 77 from both sides, we find xB=−5x_B = -5.Multiplying both sides of the second equation by 22, we get:\newline11=yB+811 = y_B + 8. Subtracting 88 from both sides, we find yB=3y_B = 3.Therefore, the coordinates of point B are (−5,3)(-5,3), which is option (B)(B).

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