# How do you write an equation for a circle tangent to the y-axis and with the center (5, 6)?

Sep 1, 2016

${x}^{2} + {y}^{2} - 10 x - 12 y + 36 = 0$

#### Explanation:

The reqd. circle touches the Y-axis [ eqn. $x = 0$] and has the Centre

at $C \left(5 , 6\right)$.

From Geometry , we know that, in the given situation, the $\bot$-

distance from the Centre to the tgt. must equal the

Radius $r$ of the circle.

$\therefore r = | 5 | = 5$.

With Centre $\left(5 , 6\right) \mathmr{and} r = 5$, the eqn. of circle is,

${\left(x - 5\right)}^{2} + {\left(y - 6\right)}^{2} = {5}^{2} - 12 y + 36 = 0$. i.e.,

graph{x^2+y^2-10x-12y+36=0 [-22.8, 22.8, -11.4, 11.42]} ${x}^{2} + {y}^{2} - 10 x - 12 y + 36 = 0$