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A circle passes through the two points (1,0) and (5,0) and is tangent…
Question
A circle passes through the two points (1,0) and (5,0) and is tangent to the line y = x. There are exactly two such circles. Find the sum of the squares of their radii.
✓ Verified answer: 46checked by our engine — not a guess
Step-by-step solution
Since the circle passes through (1,0) and (5,0), its centre is equidistant from them, so the centre lies on the perpendicular bisector x = 3.
Write the centre as (3, k) and radius r.
Through (1,0): (3-1)^2 + k^2 = r^2 => r^2 = 4 + k^2. (1)
Tangent to x - y = 0: distance from (3,k) to the line = |3 - k|/sqrt2 = r, so (3-k)^2/2 = r^2. (2)
From (1) and (2): (3-k)^2/2 = 4 + k^2 => 9 - 6k + k^2 = 8 + 2k^2 => k^2 + 6k - 1 = 0.
This gives two values of k (one for each circle), with k1 + k2 = -6 and k1 k2 = -1.
For each, r^2 = 4 + k^2. Sum of the two radii squared:
(4 + k1^2) + (4 + k2^2) = 8 + (k1^2 + k2^2) = 8 + ((k1+k2)^2 - 2 k1 k2) = 8 + (36 - 2(-1)) = 8 + 38 = 46.
Final answer46
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