How to find the equation of this circle?

[Galenus]

The circle [MATH]k[/MATH] is to be found which meets the following three conditions:

• The circle [MATH]k[/MATH] touches the x-axis in one point
• The point P(4|2)∈ [MATH]k[/MATH]
• The circle [MATH]k[/MATH] touches the circle [MATH]k_2: x^2+(y-5)^2=9[/MATH] in one point

I started with [MATH]k: (x-a)^2+(y-b)^2=R^2[/MATH],
Plugging [MATH]P[/MATH] into [MATH]k[/MATH] and noticing that [MATH]b=R [/MATH]gives the equation :
[MATH](4-a)^2+(2-R)^2=R^2[/MATH]
Next I introduced a third circle [MATH]k_3[/MATH] with the same center as [MATH]k[/MATH] and a radius of [MATH]R+3[/MATH], such that the midpoint of [MATH]k_2[/MATH] lies on [MATH]k_3[/MATH]. This gives me the equation:
[MATH]a^2+(5-R)^2=(R+3)^2[/MATH]
So I have two equation with two unknowns, but I'm stuck solving for either R or a.

 

[Dr.Peterson]

You have

[MATH](4-a)^2+(2-R)^2=R^2[/MATH]​

[MATH]a^2+(5-R)^2=(R+3)^2[/MATH]​

Try expanding both, then solve the first for R and plug that into the second. You'll get a quadratic equation, which will give two solutions. (There's actually another solution that will be tangent on the other side of the given circle, which would require changing a sign in your second equation.)