Chapter 9.2, Problem 33WE

To determine

To find: The equation of the circle.

Answer to Problem 33WE

The required center and the radius are $\left(1,-\frac{1}{4}\right)$ and $\frac{\sqrt{17}}{4}$ .

Explanation of Solution

Given information: The equation of circle is $16x^2+16y^2-32x+8y=0$ .

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

Calculation: Divide $16x^2+16y^2-32x+8y=0$ by 16.

  $x^2+y^2-2x+\frac{1}{2}y=0$

Make them complete squares by addition or subtraction,

  $\begin{array}{l}{x}^2+y^2-2x+\frac{1}{2}y=0\\ x^2-2x+1+y^2+\frac{1}{2}y+\frac{1}{16}-1-\frac{1}{16}=0\\ \left(x^2-2x+1\right)+\left(y^2+\frac{1}{2}y+\frac{1}{16}\right)-\left(-1-\frac{1}{16}\right)=0\\ \left(x^2-2x+1\right)+\left(y^2+\frac{1}{2}y+\frac{1}{16}\right)=\frac{17}{16}\\ {\left(x-1\right)}^2+{\left(y+\frac{1}{4}\right)}^2={\left(\frac{\sqrt{17}}{4}\right)}^2\end{array}$

Compare ${\left(x-1\right)}^2+{\left(y+\frac{1}{4}\right)}^2={\left(\frac{\sqrt{17}}{4}\right)}^2$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Hence, the center $\left(h,k\right)=\left(1,-\frac{1}{4}\right)$ and $r=\frac{\sqrt{17}}{4}$ .

Therefore, the center and the radius are $\left(1,-\frac{1}{4}\right)$ and $\frac{\sqrt{17}}{4}$ .