The absolute difference between the squares of the radii of the two circles passing through the point (-9, 4) and touching the lines x + y = 3 and x - y = 3, is equal to .
Answer
768
Solution
The lines meet at (3, 0), and their angle bisectors are y = 0 and x = 3.
At (-9, 4), x + y - 3 = -8 < 0 and x - y - 3 = -16 < 0, so the point lies in the angle bisected by y = 0 (the part x < 3).
Hence the centre is (h, 0) with h < 3, and the radius is its distance from x + y = 3, i.e. r = 3 - h 2.
Since the circle passes through (-9, 4),
(h + 9)2 + 16 = (3 - h)22
2h2 + 36h + 194 = h2 - 6h + 9
h2 + 42h + 185 = 0 (h + 5)(h + 37) = 0
For h = -5, r12 = 642 = 32; for h = -37, r22 = 16002 = 800.
|r22 - r12| = 800 - 32 = 768
Answer: 768