Chapter 9.3, Problem 20PPS

To determine

To Find: The equation of the circle.

Answer to Problem 20PPS

  ${(x-6)}^2+{(y+3)}^2=25$

Explanation of Solution

Given:

  

Given: $(h,k):(6,-3),(x,y):(3,1)$

The center of the circle is at $(h,k)$ and radius is at $r$

Equation of the circle

  ${(x-h)}^2+{(y-k)}^2=r^2$

Then

  $\begin{array}{}$ {(x-h)}^2+{(y-k)}^2=r^2$$ {(3-6)}^2+(1-{(-3)}^2=r^2$$ {(-3)}^2+{(4)}^2=r^2$$ r^2=9+16$$ r^2=25$\end{array}$

Substitute $r^2=25$ in circle equation

We get

  $\begin{array}{}$ {(x-h)}^2+{(y-k)}^2=r^2$$ {(x-6)}^2+{(y+3)}^2=25$\end{array}$

Hence, ${(x-6)}^2+{(y+3)}^2=25$