Chapter 9.3, Problem 51PPS

To determine

To Find: The equation for the circle that satisfies the condition.

Answer to Problem 51PPS

  ${(x-9)}^2+{(y+8)}^2=1000$

Explanation of Solution

Given: Center $(9,-8)$ , passes through $(19,22)$

  $(h,k)=(9,-8)$

Equation of the circle

  $\begin{array}{}$ {(x-h)}^2+{(y-k)}^2=r^2$$ {(x-9)}^2+{(y+8)}^2=r^2$\end{array}$

The circle is passes through $(19,22)$ so it must satisfies the equation

We get

  $\begin{array}{}$ {(x-9)}^2+{(y+8)}^2=r^2$$ {(19-9)}^2+{(22+8)}^2=r^2$$ 100+900=r^2$$ r^2=1000$\end{array}$

Substitute $r^2=1000$ in equation of the circle

  $\begin{array}{}$ {(x-h)}^2+{(y-k)}^2=r^2$$ {(x-9)}^2+{(y+8)}^2=1000$\end{array}$

Hence,

  ${(x-9)}^2+{(y+8)}^2=1000$