# How do you find the center and radius for x^2 + y^2 - 6x - 4y - 12 = 0?

Nov 1, 2016

Use the quadratic 'Complete the Square' method

#### Explanation:

x^2 - 6x +y^2 - 4y = 12

Then take 1/2 of the 'b' term for both quadratic expressions, square those values and add them to both sides.

x^2 -6x + 9 + y^2 - 4y + 4 = 12 + 9 + 4

(x - 3)^2 + (y -2)^2 = 25

Circle centered at (3,2) with radius = 5

Nov 1, 2016

Center: $\left(3 , 2\right) \textcolor{w h i t e}{\text{XXXXX}}$Radius: $5$
$\textcolor{w h i t e}{\text{XXX}}$(see below for method)

#### Explanation:

An equation in the form:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{m a \ge n t a}{r}}^{2}$
is the standard form for the equation of a circle with center $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and radius $\textcolor{m a \ge n t a}{r}$.

Lets try to convert the given equation:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + {y}^{2} - 6 x - 4 y - 12 = 0$
into the standard form for the equation of a circle.

Group the $x$ and $y$ terms separately and "move" the constant to the right side of the equation:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 6 x \textcolor{w h i t e}{+ {3}^{2}} + {y}^{2} - 4 y \textcolor{w h i t e}{+ {2}^{2}} = 12$

Complete the square for each of $x$ and $y$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 6 x \textcolor{m a \ge n t a}{+ {3}^{2}} + {y}^{2} - 4 y \textcolor{\mathmr{and} a n \ge}{+ {2}^{2}} = 12 \textcolor{m a \ge n t a}{+ {3}^{2}} \textcolor{\mathmr{and} a n \ge}{+ {2}^{2}}$

Write the left side as the sum of two squared binomials
and simplify the result on the right side
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = 25$

Express the right side as a square.
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{3}\right)}^{2} + {\left(y - \textcolor{b l u e}{2}\right)}^{2} = {\textcolor{g r e e n}{5}}^{2}$
...the equation for a circle with center $\left(\textcolor{red}{3} , \textcolor{b l u e}{2}\right)$ and radius $\textcolor{g r e e n}{5}$