Let $ABC$ be a triangle with $AB = 1$, $AC = 3$ and $\angle BAC = \pi/2$. A circle of radius $r > 0$ touches the sides $AB$ and $AC$, and also the circumcircle of $\triangle ABC$ internally. Then the value of $r$ is _____.
Step-by-Step Solution
Key Concept: Place $A$ at origin, $B = (1, 0), C = (0, 3)$. The circumcircle has centre $(1/2, 3/2)$ and radius $\sqrt{10}/2$. A circle touching both axes in the first quadrant has centre $(r, r)$. Internal tangency: $\sqrt{(r - 1/2)^2 + (r - 3/2)^2} = \sqrt{10}/2 - r$. Solve to get $r = 4 - \sqrt{10}$.
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Correct Answer: $4 - \sqrt{10}$