# How do you find the equation of the circle given radius = 4; and center = (-2, 3)?

May 27, 2016

Applying the definition of circle.

#### Explanation:

The definition of the circle is "all the point with a fixed distance from the center".
You have to translate this in a formula.

The distance between two points in the plane with coordinates $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is

$d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$.

Reading our definition of circle we want that one of the two points is always the center and the distance is the radius. Then we have

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

I used as center the first point, but in this formula you can swap the two points because the distance between two points does not change if you exchange the positions of the two points. Now, just to be more comfortable, I simply call ${x}_{2}$ as $x$ and ${y}_{2}$ as $y$. So I have

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

Just to get rid of the square root it is useful to do the square on both sides:

$16 = {\left(- 2 - x\right)}^{2} + {\left(3 - y\right)}^{2}$

and now we can expande the binomials

$16 = 4 + {x}^{2} + 4 x + 9 + {y}^{2} - 6 y$

we can reorder, move the 16 on the right side and subtract the 4 obtaining

${x}^{2} + {y}^{2} + 4 x - 6 y - 12 = 0$.