Q 35. - Precalculus Enhanced with Graphing Utilities

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.

Step-by-Step Solution

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 1. Given information.

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

Step 2. Formula used for center and radius . x 1 + x 2 2   , y 1 + y 2 2 and d = x 2 - x 1 2 + y 2 - y 1 2

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

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 - h 2 + y - k 2 = r 2

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

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