Question

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

Answer 1

Answer:

$(x-3)^{2} +(y-2)^{2}=41$

Step-by-step 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

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

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

$d=\sqrt{164}\ units$

$d=2\sqrt{41}\ units$

step 2

Find the center of the circle

The center is the midpoint of the diameter

The center is equal to

$C=(\frac{8-2}{2},\frac{-2+6}{2})$

$C=(3,2)$

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)

$r=2\sqrt{41}/2=\sqrt{41}\ units$ ---> the radius is half the diameter

substitute

$(x-3)^{2} +(y-2)^{2}=(\sqrt{41})^{2}$

$(x-3)^{2} +(y-2)^{2}=41$