Question

The endpoints of the diameter of a circle are $\left(-3,2\right)$and $\left(5,-6\right)$. Find the center and radius of the circle .Write the general equation of this circle.

Short answer

Center and radius of circle are$\left(1,-2\right)$and$\sqrt{32}$and general form of equation is$x^2+y^2-2x+4y=27$.

Step-by-step solution

Step 1. Given information.

Consider the given endpoints$\left(-3,2\right)$and$\left(5,-6\right)$.

Step 2. Formula used for center and radius .x1+x22 ,y1+y22 and d=x2-x12+y2-y12

$$\begin{gathered} \left(\frac{-3+5}{2},\frac{2-6}{2}\right) \\ \left(1,-2\right) \end{gathered}$$

Further simplify.

$$\begin{gathered} d\left(OA\right)=\sqrt{{\left(1+3\right)}^2+{\left(-2,-2\right)}^2} \\ d\left(OA\right)=\sqrt{32} \\ d\left(OB\right)=\sqrt{{\left(1-5\right)}^2+{\left(-2+6\right)}^2} \\ d\left(OB=\sqrt{32}\right) \end{gathered}$$

Further simplify.

Step 3. Using standard form of circle to find general form.x-h2+y-k2=r2

Substitute the value of $h,k,r$.

$$\begin{gathered} {\left(x-1\right)}^2+{\left(y+2\right)}^2={\sqrt{32}}^2 \\ {x}^2+y^2-2x+4y=27 \end{gathered}$$