# How do you write the equation in standard form x^2 + y^2 - 4x - 14y + 29 = 0?

Dec 1, 2015

Complete the square for $x$ and $y$ to find:

${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = 24$

or if you prefer:

${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = {\left(\sqrt{24}\right)}^{2}$

which is in standard ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ form.

#### Explanation:

$0 = {x}^{2} + {y}^{2} - 4 x - 14 y + 29$

$= \left({x}^{2} - 4 x + 4\right) + \left({y}^{2} - 14 y + 49\right) + \left(29 - 4 - 49\right)$

$= {\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} - 24$

Add $24$ to both ends to get:

${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = 24$

This is the equation of a circle with centre $\left(2 , 7\right)$ and radius $\sqrt{24} = 2 \sqrt{6}$