How do you find the center and radius of the circle x^2 + y^2 - 4x - 14y + 29 = 0?

1 Answer
Apr 15, 2018

(2,7) centre and radius $2 \sqrt{6}$

Explanation:

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

Where the centre is $\left(a , b\right)$ and the radius is $r$

So we need to use completing the square and then rearrange the equation

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

$\left[{\left(x - 2\right)}^{2} - 4\right] + \left[{\left(y - 7\right)}^{2} - 49\right] = - 29$

${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} - 53 = - 29$

${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = - 29 + 53$

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

So the centre is (2,7) and radius is $\sqrt{24}$ or $2 \sqrt{6}$