Chapter 9.3, Problem 56PPS

To determine

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

Answer to Problem 56PPS

  ${(x+2)}^2+{(y-3)}^2=4$

Explanation of Solution

Given: Center in the second quadrant, tangent to $y=1,y=5$ , and the $yaxis$

Center in the second quadrant, tangent to $y=1,y=5$

As you see from the picture

  

Equation of the circle

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

Hence, ${(x+2)}^2+{(y-3)}^2=4$