How do I find the equation of a circle given two tangents and a line which contains its centre but no points

Question

How do I find the equation of the circle which touches both the x-axis (that is y = 0 is a tangent) and the line 4x - 3y + 4 = 0, find its centre lies in the first quadrant and on the line x - y - 1 = 0.

The farthest I could go is to draw the diagrams and realize that the circle is actually uniquely determined based on the parameters given. But so far, I have not been able to get the equation.

Answer 1

Let $P(p,q)$ and $r$ be, respectively, the center of the circle and the line $4x-3y+4=0$.

We know also that $p-q-1=0$, that is, $p=q+1$.

Then $$d(OX,P)=d(r,P)$$ $$|q|=\frac{|4p-3q+4|}{\sqrt{4^2+3^2}}$$ $$5|q|=|q+8|$$

This yields two different possibilities:

• $5q=q+8\to q=2\to p=3\to P=(3,2)$
• $5q=-q-8\to q=-\frac43\to p=-\frac13\to P=\left(-\dfrac13,-\dfrac43\right)$