Question

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

Answer 1

The equation for the midpoint is given by:
 $P = (\frac{x1 + x2}{2} , \frac{y1 + y2}{2})$
 Substituting values we have:
 $P = (\frac{-10 + 4}{2} , \frac{-2 + 6}{2})$
 $P = (\frac{-6}{2} , \frac{4}{2})$
 $P = (-3 , 2)$
 We can find the radius of the circle using the equation of distance between points:
 $d = \sqrt{(x2-x1)^2 + (y2-y1)^2}$
 Substituting values:
 $d = \sqrt{(4-(-3))^2 + (6-2)^2}$
 $d = \sqrt{(4+3)^2 + (4)^2}$
 $d = \sqrt{(7)^2 + (4)^2}$
 $d = \sqrt{49 + 16}$
 $d = \sqrt{65}$
 The equation of the circle is its standard form is:
 $(x-xo) ^ 2 + (y-yo) ^ 2 = r ^ 2$
 Where,
 (xo, yo): coordinates of the center of the circle
 r: radius of the circle
 Substituting values:
 $(x + 3) ^ 2 + (y-2) ^ 2 = 65$