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Circle with Equal Intercepts, Chord √14: r² = 8 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let a circle CC 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 CC on the line x+y=1x+y=1 is 14\sqrt{14}, then the square of the radius of CC is
Solution
Answer: 8 (± 0.01)
Step 1: "Equal intercepts" with "exactly three points on the axes" means the circle touches one axis (so that axis contributes one point) and cuts the other in two — consistent with a centre (r,r)(r,r) type in the first quadrant. The perpendicular distance from the centre to the line x+y=1x+y=1 works out to 12\dfrac{1}{\sqrt2}. Step 2: For a chord of length 14\sqrt{14}, half-chord =142=\dfrac{\sqrt{14}}{2}, and by the chord relation
r2=(142)2+(r−12)2.r^2=\left(\frac{\sqrt{14}}{2}\right)^2+\left(r-\frac{1}{\sqrt2}\right)^2.
 Step 3: Expand:
r2=144+r2−2r2+12=72+r2−2 r+12.r^2=\frac{14}{4}+r^2-\frac{2r}{\sqrt2}+\frac12=\frac72+r^2-\sqrt2\,r+\frac12.
 Step 4: Cancel r2r^2 and solve:
0=72+12−2 r ⇒ 2 r=4 ⇒ r=22.0=\frac72+\frac12-\sqrt2\,r\ \Rightarrow\ \sqrt2\,r=4\ \Rightarrow\ r=2\sqrt2.
 Step 5:
r2=(22)2=8.r^2=(2\sqrt2)^2=8.
 Correct answer: 8
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