Question 1040187
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Find the equation of the circle which is tangent to the x-axis at (2,0) and touches the line y=6.
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<pre>
Notice that y= 0 and y = 6 are two parallel tangent lines.
So the diameter of the circle is 6 and the radius is 3.

The center of the circle is on the vertical line x = 2.

Hence, the center is at the point (2,3).

So the equation of the circle is 

{{{(x-2)^2 + (y-3)^2}}} = {{{3^2}}},   or

{{{(x-2)^2 + (y-3)^2}}} = {{{9}}}.
</pre>

Solved.