Question

Write the equation of the circle whose diameter has endpoints (-16, -16) and (-4, -8). *

Answer 1

Answer:

The equation of circle is $(x+10)^{2}$ + $(y+12)^{2}$ = 52

Step-by-step explanation:

Given the endpoints of diameter of a circle: (-16,-16) and (-4,-8)

We know that the equation of circle is given by

$(x-h)^{2}$ + $(y-k)^{2}$ = $r^{2}$

where (x,y) is any point on the circle, (h,k) is center of the circle and r is radius of circle.

To find (h,k): the center is midpoint of diameter

Midpoint of diameter with end points (x1,y1) and (x2,y2) is given by

(  $\frac{x1+x2}{2}$ , $\frac{y1+y2}{2}$ )

(  $\frac{-16-4}{2}$ , $\frac{-16-8}{2}$ )

(-10, -12)

Hence (h,k) is (-10,-12)

Substituting values of (h.k) and (x.y) as (-10,-12) and (-4,-8) respectively in equation of circle, we get

$(-4+10)^{2}$ + $(-8+12)^{2}$ = $r^{2}$

$r^{2}$ = 52

Substituting values of (h.k) and $r^{2}$, we get the equation of circle as

$(x+10)^{2}$ + $(y+12)^{2}$ = 52

Hence the equation of circle is $(x+10)^{2}$ + $(y+12)^{2}$ = 52