# What is the center and radius of the circle with equation x^2 + y^2 - 2x + 14y + 42 = 0?

Jun 18, 2018

the centre is $\left(1 , - 7\right)$ and the radius is $\sqrt{8} = 2 \sqrt{2}$

#### Explanation:

${x}^{2} + {y}^{2} - 2 x + 14 y + 42 = 0$

$\left({x}^{2} - 2 x\right) + \left({y}^{2} + 14 y\right) = - 42$

Complete the square. If you have an equation $a {x}^{2} + b x$ and you want to complete the square, all you have to do is to half the coefficient of the term $x$ ie $\frac{b}{2}$ and then square it ie ${\left(\frac{b}{2}\right)}^{2}$. Your equation will end up being $a {x}^{2} + b x + {\left(\frac{b}{2}\right)}^{2}$

$\left({x}^{2} - 2 x + 1\right) + \left({y}^{2} + 14 y + 49\right) = - 42 + 1 + 49$

${\left(x - 1\right)}^{2} + {\left(y + 7\right)}^{2} = 8$

Since the general form of the circle is given by: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ where the centre is $\left(h , k\right)$ and the radius is r, then looking at the above equation, the centre is $\left(1 , - 7\right)$ and the radius is $\sqrt{8} = 2 \sqrt{2}$