# How do you write the standard form of the equation of the circle that is tangent to x=-2 and has its center at (2, -4)?

Jun 14, 2016

Equation of circle is ${x}^{2} + {y}^{2} - 4 x + 8 y + 4 = 0$

#### Explanation:

Distance from a point $\left(m , n\right)$ to a line $a x + b y + c = 0$ is given by

$\frac{a m + b n + c}{\sqrt{{a}^{2} + {b}^{2}}}$

Hence, length of perpendicular from $\left(2 , - 4\right)$ to $x = - 2$ or $x + 2 = 0$ is $\frac{2 + 2}{1} = 4$

Hence $4$ is radius and as center is $\left(2 , - 4\right)$, equation of circle will be

${\left(x - 2\right)}^{2} + {\left(y + 4\right)}^{2} = 16$ or

${x}^{2} - 4 x + 4 + {y}^{2} + 8 y + 16 = 16$

or ${x}^{2} + {y}^{2} - 4 x + 8 y + 4 = 0$