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?
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?
1 Answer
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
The radius at the tangent point
Our circle equation so far is
We plug in our one known point to solve for
So the equation is
graph{(x - 1)^2 + (y- 13/3)^2 = (13/3)^2 [-8.87, 11.13, -0.88, 9.12]}