The tangent to $x^2 + y^2 = 25$ at $R(3, 4)$ meets the $x$-axis at $P$ and the $y$-axis at $Q$. A circle through the origin $O$ has its centre at the incentre of $\triangle OPQ$. If $r$ is its radius, then $r^2 =$
Step-by-Step Solution
Key Concept: Tangent: $3x + 4y = 25$. So $P = (\frac{25}{3}, 0)$, $Q = (0, \frac{25}{4})$. Compute $OP, OQ, PQ$ and find incentre $I = (\frac{25}{12}, \frac{25}{12})$. Then $r^2 = |OI|^2 = 2 \cdot (\frac{25}{12})^2 = \frac{625}{72}$.
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Correct Answer: (1)