Question

If the endpoints of the diameter of a circle are (−8, 0) and (−12, 2), what is the standard form equation of the circle?

Answer 1

Answer:

The standard form equation of circle is

$(x+10)^{2}$ +$(y-1)^{2}$ = 5

Step-by-step explanation:

Given the endpoints of diameter of circle (-8,0) and (-12,2)

The equation of circle is given by

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

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

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

Midpoint of diameter with endpoints (x1,y1) and (x2,y2) is

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

Substituting the values of endpoints

(  $\frac{-8-12}{2}$ , $\frac{0+2}{2}$ )

(-10,1)

(h,k) is (-10,1)

Substituting the values of (h,k) and (x,y) as (-10,1) and (-8,0) respectively in equation of  circle we get,

$(-8+10)^{2}$ + $(0-1)^{2}$ = $r^{2}$

$r^{2}$ = 5

Now substituting the values  of (h,k) and $r^{2}$ , we get

The standard form equation of circle as

$(x+10)^{2}$ +$(y-1)^{2}$ = 5