Circlomics

In order for the circle to be tangent to the two axes at described, the center of the circle must lie on the line y = -x.

The radius is then equal to y. This must be the same distance from the center to the point (-8,9).

y = sqrt( [x - (-8)]^2 + [y-9]^2)

y = sqrt( [x+8]^2 + [y-9]^2)

and since y = -x,

y = sqrt ([-y+8]^2 + [y-9]^2)

Solving for y yields y = 5 or 29. The sum of which is 34.

Since the radii are a positive numbers the easy way to tackle the problem is assume our circumference located in first quadrant and mirror the coordinates of the given point to M (a,b). By the way this two circumferences must also pass over point M’ (b,a) (symmetrical to M with respect of bisector of first quadrant), but we will not need to use it for now.

The equations of circumferences with center in the mentioned bisector and tangent to main axis will be :

(x-r)^2 + (y-r)^2 = r^2 once expanded we get r^2 - 2r (x+y) + x^2 + y^2 = 0 Because our circumference must pass on M by substitution of coordinates of M(a,b) we will obtain r^2 - 2r (a+b) + a^2 + b^2 = 0. Since the problem ask for the sum r1+r2 the two roots of quadratic equation we know is the coefficient of r, once sign is changed, its value I therefore 2 ( 8+9) = 34

The general formula for a circle is $(x-a)^2 + (y-b)^2 = r^2$ where $(a, b)$ is the centre.

For the circle to be tangent to both $x$ and $y$ , $a, b = r$ . Also note that the circles are tangent to the negative $x$ -axis so the we use is actually, $(x+r)^2 + (y-r)^2 = r^2$ . Therefore setting $(x, y) = (-8, 9)$ and expanding gives us $r^2 - 16r + r^2 - 18r + 145 = r^2 \implies r^2 - 34r +145 = 0$ . Solving the quadratic we get that $r = 17 \pm 12 \therefore r= 5, 29$ and $29 + 5 = \boxed{34}$ .