Question

find an equation of the circle whose diameter has endpoints (6, -3) and (-2, 3).

Explanation:

Step1: Find the center of the circle

The center of the circle is the mid - point of the diameter. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.
For $(x_1,y_1)=(6,-3)$ and $(x_2,y_2)=(-2,3)$, the center $(h,k)$ is $(\frac{6+( - 2)}{2},\frac{-3 + 3}{2})=(2,0)$.

Step2: Find the radius of the circle

The radius $r$ is the distance from the center $(h,k)$ to either of the endpoints of the diameter. The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
Using the center $(h,k)=(2,0)$ and the point $(x_1,y_1)=(6,-3)$, we have $r=\sqrt{(6 - 2)^2+(-3 - 0)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.

Step3: Write the equation of the circle

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$.
Substituting $h = 2$, $k = 0$, and $r = 5$ into the equation, we get $(x - 2)^2+(y-0)^2=25$, which simplifies to $(x - 2)^2+y^2=25$.

Answer:

$(x - 2)^2+y^2=25$