`(x-6)^2 + (y+5)^2 =16`
In the xy-plane, the graph of the equation above is a circle. Point P is on the circle and has coordinates (10, −5). If `bar(PQ)` is a diameter of the circle, what are the coordinates of point Q ?

Options

A. (2, −5)
B. (6, −1)
C. (6, −5)
D. (6, −9)

Answer

A

Explanation

The standard form for the equation of a circle is `(x – h)^2 + (y – k)^2 = r^2`, where (h, k) are the coordinates of the center and r is the length of the radius. According to the given equation, the center of the circle is (6, –5). Let `(x_1, y_1)` represent the coordinates of point Q. Since point P (10, –5) and point Q `(x_1, y_1)` are the endpoints of a diameter of the circle, the center (6, –5) lies on the diameter, halfway between P and Q. Therefore, the following relationships hold: `(x_1+10)/2=6` and `(y_1+(-5))/2=-5`. Solving the equations for `x_1` and `y_1`, respectively, yields `x_1 = 2` and `y_1` = −5. Therefore, the coordinates of point Q are (2, –5)
Alternate approach: Since point P (10, −5) on the circle and the center of the circle (6, −5) have the same y-coordinate, it follows that the radius of the circle is 10 – 6 = 4. In addition, the opposite end of the diameter `bar(PQ)` must have the same y-coordinate as P and be 4 units away from the center. Hence, the coordinates of point Q must be (2, –5).
Choices B and D are incorrect because the points given in these choices lie on a diameter that is perpendicular to the diameter `bar(PQ)` . If either of these points were point Q, then `bar(PQ)` would not be the diameter of the circle. Choice C is incorrect because (6, −5) is the center of the circle and does not lie on the circle.