Mathematics · Circles

JEE Main 2026 — 6 April, Morning Shift — Question 42

Let the centre of the circle $x²+y²+2gx+2fy+25=0$ be in the first quadrant and lie on the line $2x-y=4.$ Let the area of an equilateral triangle inscribed in the circle be $27\sqrt3.$ Then the square of the length of the chord of the circle on the line $x=1$ is _____.

Answer: 80

Numerical answer — enter this value.

Step-by-step solution

$x^{2}+y^{2}+2 g x+2 f y+25=0$ $\mathrm{C} \equiv(-\mathrm{g},-\mathrm{f})$ lies on line

\therefore-2 \mathrm{~g}+\mathrm{f}=4 \end{gathered}$$ Also $\mathrm{g}^{2}+\mathrm{f}^{2}-25=\mathrm{r}^{2}$ Area of equation triangle inside the circle $=\frac{3 \sqrt{3}}{4} \mathrm{r}^{2}$ $27 \sqrt{3}=\frac{3 \sqrt{3}}{4} \times \mathrm{r}^{2}$ $\mathrm{r}^{2}=36$ Solving (i) and (ii) $\mathrm{g}=-5, \mathrm{f}=-6$ $\ell_{\mathrm{AB}}=\sqrt{36-16}=2 \sqrt{20}$ $\left(\ell_{\mathrm{AB}}\right)^{2}=80$

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: Circles
Topic: Introduction to Circles