Question

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

Answer 1

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^2-6x-12y+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 `(3,6)` to the X-axis
`=`y-co-ord. of `C(3,6)=6`.

Accordingly, the eqn. of the circle is `: (x-3)^2+(y-6)^2=6^2`, i.e.,

`x^2+y^2-6x-12y+9+0`