Chapter 11, Problem 8CR

To determine

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

Answer to Problem 8CR

Solution:

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

2) The graph of the circle $x^2+y^2-2x+4y-11=0$ is,

Explanation of Solution

Given information:

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

Explanation:

The standard form of an equation of a circle with radius $r$ and center $\left(h,k\right)$ is

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

The given equation is $x^2+y^2-2x+4y-11=0$.

Combine the same variable terms,

$\left(x^2-2x\right)+\left(y^2+4y\right)-11=0$

To complete the square, add and subtract the number to each bracket,

$\Rightarrow \left(x^2-2x+1-1\right)+\left(y^2+4y+4-4\right)-11=0$

By simplifying,

$\Rightarrow \left(x^2-2x+1\right)+\left(y^2+4y+4\right)-11-1-4=0$

Now by factoring,

$\Rightarrow {\left(x-1\right)}^2+{\left(y+2\right)}^2-16=0$

$\Rightarrow {\left(x-1\right)}^2+{\left(y+2\right)}^2=16$

$\Rightarrow {\left(x-1\right)}^2+{\left(y+2\right)}^2=4^2$

Compare the above equation with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

$\Rightarrow h=1,k=-2,r=4$.

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

The graph of the circle $x^2+y^2-2x+4y-11=0$ is ,