Q. 8 - Precalculus Enhanced with Graphing Utilities

Question

Find the center and radius of the circle $x^2+y^2-2x+4y-11=0$. Graph the circle.

Step-by-Step Solution

Answer

The center of the circle $x^2+y^2-2x+4y-11=0$ is, $\left(1,-2\right)$ and the radius is, $4$ units.

The graph of the circle is,

Step 1 Given equation is,

$x^2+y^2-2x+4y-11=0$.

The equation of the circle is, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$

where $\left(h,k\right)$ is the center and $a$ is the radius.

Now simplify the given equation to the form of the circle.

$$\begin{gathered} x^2+y^2-2x+4y-11=0 \\ {x}^2-2x+y^2+4y=11 \\ {x}^2-2x+{\left(1\right)}^2+y^2+4y+{\left(2\right)}^2=11+{\left(1\right)}^2+{\left(2\right)}^2 \\ {\left(x-1\right)}^2+{\left(y+2\right)}^2=11+1+4 \\ {\left(x-1\right)}^2+{\left(y+2\right)}^2=16 \\ {\left(x-1\right)}^2+{\left(y+2\right)}^2={\left(4\right)}^2 \end{gathered}$$

Step 2 Now compare the obtained equation with the equation of the circle.

We get, the center is,$\left(1,-2\right)$ and the radius is, $4$.

Now let us draw a graph for the given equation of the circle whose center is marked by the point $C$.