Question

Find the equation of a circle passing through (5,6) and tangent to the x-axis at (1,0)?

I am completely lost with this one. I know the tangent will be on the line x=1, but does the actual point of intersection of the tangent and the circle have an x-coordinate of 1? How do I get the y-coordinate?

Answer 1

`(x - 1)^2 + (y- 13/3)^2 = (13/3)^2`

Explanation:

When a circle is tangent to the x axis it's either completely above it or completely below it. Here we know there's a point in the first quadrant, so the circle is above the `x` axis.

The radius at the tangent point `(1,0)` must be perpendicular to the `x` axis, i.e. vertical. So if the radius is length `r` then the center coordinates must be `(1,r)`.

Our circle equation so far is

`(x - 1)^2 + (y-r)^2 = r^2`

We plug in our one known point to solve for `r`.

`(5-1)^2 + ( 6-r)^2 = r^2`

`16 + 36 - 12r + r^2 = r^2`

`52 = 12r`

`r = 13/3`

So the equation is

`(x - 1)^2 + (y- 13/3)^2 = (13/3)^2`

graph{(x - 1)^2 + (y- 13/3)^2 = (13/3)^2 [-8.87, 11.13, -0.88, 9.12]}