Question

What is the equation of the circle whose diameter has endpoints (8,-2) and (-2,6)?

Answer 1

`(x-3)^(2) +(y-2)^(2)=41`

Explanation:

step 1

Find the diameter of the circle

the formula to calculate the distance between two points is equal to

`d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}`

we have

`A(8,-2)\\E(-2,6)`

substitute the values

`d=\sqrt{(6+2)^(2)+(-2-8)^(2)}`

`d=\sqrt{(8)^(2)+(-10)^(2)}`

`d=√(164)\ units`

`d=2√(41)\ units`

step 2

Find the center of the circle

The center is the midpoint of the diameter

The center is equal to

`C=((8-2)/(2),(-2+6)/(2))`

`C=(3,2)`

step 3

Find the equation of the circle

The equation of the circle in center radius form is equal to

`(x-h)^(2) +(y-k)^(2)=r^(2)`

we have

(h,k)=(3,2)

`r=2√(41)/2=√(41)\ units` ---> the radius is half the diameter

substitute

`(x-3)^(2) +(y-2)^(2)=(√(41))^(2)`

`(x-3)^(2) +(y-2)^(2)=41`