Question

Find the center and the radius of the circle$x^2+4x+y^2-8y=16$.

(Hint: Express the given equation in the form

${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$)

Short answer

The center is (-2, 4) and the radius is 6.

Step-by-step solution

Step-1 – Given

The given equation is: $x^2+4x+y^2-8y=16$.

Step-2 – To determine

We have to find the center and radius of the given circle.

Step-3 – Calculation 

We first write the equation to the standard form of a circle:

$\begin{array}{l}{x}^2+4x+y^2-8y=16\\ x^2+4x+4+y^2-8y+16=16+20\left[\text{addbothsides20}\right]\\ \left[\left(x^2\right)+2\left(2\right)\left(x\right)+{\left(2\right)}^2\right]+\left[{\left(y\right)}^2-2\left(4\right)\left(y\right)+{\left(4\right)}^2\right]=36\left[\text{perfectsquaretrinomial}\right]\\ {\left(x+2\right)}^2+{\left(y-4\right)}^2={\left(6\right)}^2\end{array}$

The standard form of a circle is:${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

Where (a, b) = center and r = radius.

Then we compare the given equation to the standard form of a circle:

$a=-2,b=4\text{and}r=6$.

So, the center is (-2, 4) and the radius is 6.