# How do you write an equation with center (3,6), tangent to the x-axis?

Jul 18, 2016

The Question is not clear, but, I presume that what is reqd. is the eqn. of a circle having centre at (3,6) and the X-axis is tgt. to the reqd. circle.

${x}^{2} + {y}^{2} - 6 x - 12 y + 9 + 0$

#### Explanation:

The Question is not clear, but, I presume that what is reqd. is the eqn. of a circle having centre at (3,6) and the X-axis is tgt. to the reqd. circle.

In that case, we know from Geometry that the $\bot$ dist. from the centre to a tgt. of a circle is equal to the radius of the circle.

Hence, $r =$radius$= \bot$ dist. from centre $\left(3 , 6\right)$ to the X-axis
$=$y-co-ord. of $C \left(3 , 6\right) = 6$.

Accordingly, the eqn. of the circle is $: {\left(x - 3\right)}^{2} + {\left(y - 6\right)}^{2} = {6}^{2}$, i.e.,

${x}^{2} + {y}^{2} - 6 x - 12 y + 9 + 0$