Chapter 9.2, Problem 31WE

To determine

To find: The equation of the circle.

Answer to Problem 31WE

The required center and the radius are $\left(2,3\right)$ and $2$ .

Explanation of Solution

Given information: The equation of circle is $4x^2+4y^2-16x-24y+36=0$ .

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

Calculation: Divide $4x^2+4y^2-16x-24y+36=0$ by 4.

  $x^2+y^2-4x-6y+9=0$

Make them complete squares by addition or subtraction,

  $\begin{array}{l}{x}^2-4x+y^2-6y+9=0\\ x^2-4x+4+y^2-6y+9-4=0\\ \left(x^2-4x+4\right)+\left(y^2-6y+9\right)=4\\ {\left(x-2\right)}^2+{\left(y-3\right)}^2=2^2\end{array}$

Compare ${\left(x-2\right)}^2+{\left(y-3\right)}^2=2^2$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Hence, the center $\left(h,k\right)=\left(2,3\right)$ and $r=2$ .

Therefore, the center and the radius are $\left(2,3\right)$ and $2$ .