# How do you find the center and radius of the circle given x^2+(y-sqrt3)^2+4x=25?

Nov 18, 2016

The center is $\left(- 2 , \sqrt{3}\right)$ and the radius is $\sqrt{29}$

#### Explanation:

The standard form for the equation of a circle is:

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

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center, and r is the radius.

Please notice that (y - sqrt(3))^2 is already in that form and we can see that $k = \sqrt{3}$.

Add ${h}^{2}$ to both sides of the given equation:

${x}^{2} + 4 x + {h}^{2} + {\left(y - \sqrt{3}\right)}^{2} = 25 + {h}^{2}$

When we expand the x square in the standard form we get:

${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$

We can find the value of h by setting the middle term from the standard form equal to the middle term of our equation:

$- 2 h x = 4 x$

$h = - 2$

This means that we can substitute ${\left(x - - 2\right)}^{2}$ for the terms on the left and 4 for ${h}^{2}$ on the right:

${\left(x - - 2\right)}^{2} + {\left(y - \sqrt{3}\right)}^{2} = 25 + 4$

Combine the right side:

${\left(x - - 2\right)}^{2} + {\left(y - \sqrt{3}\right)}^{2} = 29$

Write the 29 as ${\left(\sqrt{29}\right)}^{2}$

${\left(x - - 2\right)}^{2} + {\left(y - \sqrt{3}\right)}^{2} = {\left(\sqrt{29}\right)}^{2}$

In this form we can see the center and radius by observation:

The center is $\left(- 2 , \sqrt{3}\right)$ and the radius is $\sqrt{29}$