JEE Main 2026 — Apr 2, Shift 1 — Mathematics solution

JEE Main2026Apr 2, Shift 1Mathematics

Let a circle $C$ have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of $C$ on the line $x + y = 1$ is $14$ , then the square of the radius of $C$ is _____.

Your answer

Solution

$r^2 = \left(\dfrac{\sqrt{14}}{2}\right)^2 + \left(r-\dfrac{1}{\sqrt{2}}\right)^2$

$r^2 = \dfrac{7}{2} + r^2 - \sqrt{2}\,r + \dfrac{1}{2}$

$2 r = 4$

$r = 22$

$r^{2} = 8$

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