Find the equation of a circle in the 3rd quadrant that is tangent to the line y=x and the x-axis, with a radius of 5.
One way I thought of doing it was letting the center point of the circle be the point (-x, -5) and the point of tangency between the circle and y=x be the point (-a, -a). Then using the distance formula and the slope formula between the points, you have two equation in two variables which you can then solve.
However, this method is tedious. I am wondering if there is a slicker, more elegant solution that is more geometric in nature and uses the angles and side lengths given.