How do I find the equation of the corresponding circle?

Question

Let $C$ be a circle whose center belongs to first quadrant. Circle $C$ is tangent to lines $y + 3 = 0$ and $4y - 3x - 12 = 0$. Given information what is the equation for circle $C$.

My Attempt:

I found that center coordinates $O(x_1 , y_1)$ has relation $3y_1 = x_1 - 1$ and $r=| y_1 + 3| = ( |4y_1 - 3x_1 -12| ) /5$

Answer 1

There are infinitely many circles can be drawn touching these two lines.

The characteristics of these circle are having their centers lying on the angle bisector between these two lines.

The equation of that angle bisector can by found either by formula; or

the following hard way:-

-(1) find the point of intersection of these two lines;

-(2) find the angle between these two lines;

-(3) use half the angle found in (2) to get the slope;

-(4) use point-slope form to build the equation of that angle bisector.

That angle bisector happens to be -x +3y = -1.

Next, we have to find where the line will cut the x-axis. [Answer (1, 0)].

Conclusion: There are infinitely many circles can be drawn touching these two lines. The centers of these circles ly on the line ... and the smallest one is centered at .... with radius = ...

Answer 2

You can assume the equation of a circle $$ (x-x_0)^2 + (y-y_0)^2 = r^2 $$ and try to determine the three unknowns $(x_0, y_0)$ and $r$ from the other information.