Chapter 9.2, Problem 39WE

To determine

To find: The equation of the circle.

Answer to Problem 39WE

The required equation is $x^2+y^2+4x-8y+4=0$ .

Explanation of Solution

Given information: The center on line $y=4$ and tangent to x-axis at $\left(-2,0\right)$ .

Formula used: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: As we know that the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ has the center $\left(h,k\right)$ and radius $\left(r\right)$ .

  

Since the center is -2 units left of $\left(0,4\right)$ . Hence, the center is $\left(h,k\right)=\left(-2,4\right)$ .

By the above sketch, the radius $r=4$

Now, substitute the value of $\left(h,k\right)$ and $\left(r\right)$ into the equation,

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-\left(-2\right)\right)}^2+{\left(y-4\right)}^2=4^2\\ {\left(x+2\right)}^2+{\left(y-4\right)}^2=16\\ x^2+4x+4+y^2-8y=16-16\\ x^2+y^2+4x-8y+4=0\end{array}$

Therefore, the required equation is $x^2+y^2+4x-8y+4=0$ .

The graph is given below,