Question

PQ is the diameter of the circle whose centre is m(3,-2) and q(4,-6). find the coordinates of p​

Answer 1

Answer:

$center \: of \: the \: circle = middle \: of \: pq \\ = ( \frac{x2 - x1}{2} \: \frac{y2 - y1}{2} )$

(3,-2). = (4-x/2 ,-6+ y/2)

4-x/2=3 -6 +y/2 = -2

4-x=6 -6 -y = -4

x = 4-6 y=-6+4

x = -2. y=-2

p =(-2,-2)