Chapter 9, Problem 3CT

To determine

To find Center and radius of a Circle - $x^2+y^2-8x+2y+11=0$

Answer to Problem 3CT

The Center is $\left(4,-1\right)$ and radius is $\sqrt{6}$ .

Explanation of Solution

Given:

Circle - $x^2+y^2-8x+2y+11=0$

Formula Used:

Circle equation is given as $x^2+y^2+2gx+2fy+c=0$ ; where $\left(-g,-f\right)$ is the Center and $r=\sqrt{g^2+f^2-c}$ is the radius.

Calculation:

Given:

Circle - $x^2+y^2-8x+2y+11=0$

Using the above formula and comparing the above values in the equation of the Circle, we have:

  $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2-8x+2y+11=0$

Thus,

  $\begin{array}{l}2g=-8\\ g=-4\end{array}$

  $\begin{array}{l}2f=2\\ f=1\end{array}$

  $c=11$

Thus, Center of the Circles is $\left(-g,-f\right)$

Center of the Circles $=$$\left(4,-1\right)$

Now, Radius:

  $\begin{array}{l}r=\sqrt{g^2+f^2-c}\\ r=\sqrt{4^2+{\left(-1\right)}^2-11}\\ r=\sqrt{16+1-11}\\ r=\sqrt{6}\end{array}$

Thus, radius is $\sqrt{6}$ .

Thus, the Center is $\left(4,-1\right)$ and radius is $\sqrt{6}$ .

Conclusion:

The Center is $\left(4,-1\right)$ and radius is $\sqrt{6}$ .