SOLUTION: find the equation of circle which touches xaxis at (4,0) and also passes through (2,2)

Question 1106993: find the equation of circle which touches xaxis at (4,0) and also passes through (2,2)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Let's label the two points as
A = (4,0)
B = (2,2)

The circle is tangent to the x axis at point A since it only touches the x axis at this point. The center C will lay on a line perpendicular to the x axis, and this line will go through point A. Call this line L

Line L is the equation as this is a vertical line perpendicular to the horizontal line that makes up the x axis. This vertical line goes through (4,0). All points on the line have x coordinate of 4 (y can be anything you want)

Let's find the perpendicular bisector of line segment AB. Call this line M. The center will lay on the perpendicular bisector of these two points (A and B). It turns out the slope of AB is -1 after using the slope formula. So anything perpendicular to this has slope of m = 1 (use negative reciprocal rule). Plugging m = 1, x = 2, y = 2 into y = mx+b and solving for b leads to b = -2. So the equation of line M is y = 1x-2 or simply y = x-2

To recap:
Equation of line L: x = 4
Equation of line M: y = x-2

Solve this system. To do so, we use substitution. Replace x with 4 and simplify
y = x-2
y = 4-2
y = 2
The point C is the result of intersecting lines L and M
The two lines cross at (x,y) = (4,2)

So point C is C = (4,2)
This is the center of the circle
Center = (h,k) = (4,2)

The radius is the distance from either C to A, or the distance from C to B. The easiest to compute is the distance from C to A, which is 2 units. Simply count the spaces between the two points.
Therefore the radius is r = 2

Plug (h,k) = (4,2) and r = 2 into the equation below

The final answer is

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