# How do you find the equation of a circle center is at (-2, 3) and that is tangent to the line 20x - 21y - 42 = 0?

##### 1 Answer
Mar 31, 2017

${\left(x + 2\right)}^{2} + {\left(y - 3\right)}^{2} = {5}^{2} ,$

or,

${x}^{2} + {y}^{2} + 4 x - 6 y - 12 = 0.$

#### Explanation:

Let us name the Centre $C ,$ Radius $r$ and, the Tgt. Line $t .$

$\therefore C = C \left(- 2 , 3\right) , \mathmr{and} , t : 20 x - 21 y - 42 = 0.$

From Geometry, we know that, the $\bot -$dist. from $C$ to $t$ is $r$.

Recall that the $\bot -$dist. from pt. $\left(h , k\right)$ to line $l : a x + b y + c = 0$

is $| a h + b k + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}}$.

$\therefore r = | 20 \left(- 2\right) - 21 \left(3\right) - 42 \frac{|}{\sqrt{{20}^{2} + {\left(- 21\right)}^{2}}} = \frac{145}{29} = 5.$

Hence follows the Eqn. of the Circle, ${\left(x + 2\right)}^{2} + {\left(y - 3\right)}^{2} = {5}^{2} ,$ or,

${x}^{2} + {y}^{2} + 4 x - 6 y - 12 = 0.$

Enjoy Maths.!