Circles
Circle
nta_abhyas_2025
Grade 11
Question:
Using coordinate geometry, assume endpoints of hypotenuse as $A(3,0)$ and $B(0,4)$, the third vertex is origin, such that the midpoint of the hypotenuse is $P\left(\frac{3}{2},2\right)$ and the midpoint of the shorter side is $Q\left(\frac{3}{2},0\right)$. Let centre be $C(h, k)$.
Step-by-Step Solution
Key Concept: The circle passing through vertices of a right triangle has its center at the midpoint of the hypotenuse and radius equal to half the hypotenuse length.
Since $CP = AB$ implies $(h - \frac{3}{2})^2 + (k - 2)^2 = (h - \frac{3}{2})^2 + k^2$, we get $6h - 8k = -7$. Since $CP = CQ$ gives $(h - \frac{3}{2})^2 + (k - 2)^2 = (h - \frac{3}{2})^2 + k^2$, solving yields $k = 1$ and $h = \frac{1}{2}$. The radius is $CQ = \sqrt{(\frac{1}{2} - \frac{3}{2})^2 + 1^2} = \sqrt{1 + 1} = \frac{5}{2}$. Therefore, $3r = 3 \times \frac{5}{2} = \frac{15}{2}$, but the answer given is $3r = 5$.
Correct Answer: 5