Chapter 11.5, Problem 11MRE

To determine

To calculate:

Radius of the circle $x^2+y^2-4x+6y-3=0$ .

Answer to Problem 11MRE

Radius of the circle $x^2+y^2-4x+6y-3=0$ is $4$ .

Explanation of Solution

Give information:

Expression:

  $x^2+y^2-4x+6y-3=0$

Calculation:

Given is,

  $x^2+y^2-4x+6y-3=0$

Transform he given equation in ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ by completing the square relative to $x$ and relative to $y$ .

In order to complete the square,

  $\begin{array}{l}{x}^2-4x+4+y^2+6y+9=3+4+9\\ {\left(x-2\right)}^2+{\left(y+3\right)}^2=16\end{array}$

Compare the equation ${\left(x-2\right)}^2+{\left(y+3\right)}^2=16$ and ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Radius of the circle is given by,

  $\begin{array}{l}{r}^2=16\\ r=\sqrt{16}\\ r=4\end{array}$

Therefore, Radius of the circle $x^2+y^2-4x+6y-3=0$ is $4$ .