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

1 Answer
Jul 18, 2016

Answer:

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#