# Equation of circle tangent to y=x and x-axis with radius 5

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.

If the coordinate of the center is $(-b,-5)$ as it lies on $y=-5$
Observe that $$\tan 22.5^\circ=-\frac5b$$
• @user2612743, then use $$(b-0)^2+(0-0)^2=(a-0)^2+(a-0)^2$$ which is the square of the length of the tangent and $$(-5-a)^2+(-b-a)^2=5^2$$ which the square of the radius – lab bhattacharjee Jan 20 '14 at 4:12
I don't know if this is any slicker. Draw in one more tangent line segment from $y=x$ to the $x$-axis creating an isosceles right triangle with an inscribed circle. Call the length of the segment from the origin to the circle $a$ (2 line segments tangent to a circle from the same point have the same length by theorem, construct a line segment from the origin to the center of the circle to prove it in this instance). Each leg has length $a+5$ while the hypotenuse has length $2a$. Use the Pythagorean Theorem to solve.