Q.22. Let the centre of the circle be in the first quadrant and lie on the line . Let the area of a

Q: What is the correct answer for JEE Main Mathematics Q22?

The correct answer is 80. Sol. The center of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ is $C(-g, -f)$. It lies in the first quadrant, so $-g > 0 \Rightarrow g 0 \Rightarrow f < 0$. The center lies on $2x - y = 4$, so $2(-g) - (-f) = 4 \Rightarrow -2g + f = 4$ (i). The radius of the circle is $r = \sqrt{g^2 + f^2 - 25}$. For an equilateral triangle inscribed in a circle, its side length $a = r\sqrt{3}$. The

Q: Q.22. Let the centre of the circle be in the first quadrant and lie on the line . Let the area of an equilateral triangle inscribed in the circle be . Then the square of the length of the chord of…

The correct answer is 80. Sol. The center of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ is $C(-g, -f)$. It lies in the first quadrant, so $-g > 0 \Rightarrow g 0 \Rightarrow f < 0$. The center lies on $2x - y = 4$, so $2(-g) - (-f) = 4 \Rightarrow -2g + f = 4$ (i). The radius of the circle is $r = \sqrt{g^2 + f^2 - 25}$. For an equilateral triangle inscribed in a circle, its side length $a = r\sqrt{3}$. The

Question

Q.22. Let the centre of the circle  be in the first quadrant and lie on the line . Let the area of an equilateral triangle inscribed in the circle be . Then the square of the length of the chord of the circle on the line  is ________.

FirstInTest · Accepted Answer

Sol. The center of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ is $C(-g, -f)$. It lies in the first quadrant, so $-g > 0 \Rightarrow g < 0$ and $-f > 0 \Rightarrow f < 0$. The center lies on $2x - y = 4$, so $2(-g) - (-f) = 4 \Rightarrow -2g + f = 4$ (i). The radius of the circle is $r = \sqrt{g^2 + f^2 - 25}$. For an equilateral triangle inscribed in a circle, its side length $a = r\sqrt{3}$. The area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (r\sqrt{3})^2 = \frac{\sqrt{3}}{4} (3r^2) = \frac{3\sqrt{3}}{4} r^2$. Given area is $27\sqrt{3}$. So, $\frac{3\sqrt{3}}{4} r^2 = 27\sqrt{3} \Rightarrow \frac{3}{4} r^2 = 27 \Rightarrow r^2 = 36$. Thus, $g^2 + f^2 - 25 = 36 \Rightarrow g^2 + f^2 = 61$ (ii). From (i), $f = 4 + 2g$. Substitute into (ii): $g^2 + (4+2g)^2 = 61 \Rightarrow g^2 + 16 + 16g + 4g^2 = 61 \Rightarrow 5g^2 + 16g - 45 = 0$. Using the quadratic formula, $g = \frac{-16 \pm \sqrt{16^2 - 4(5)(-45)}}{2(5)} = \frac{-16 \pm \sqrt{256 + 900}}{10} = \frac{-16 \pm \sqrt{1156}}{10} = \frac{-16 \pm 34}{10}$. So $g = \frac{18}{10} = 1.8$ or $g = \frac{-50}{10} = -5$. Since $g < 0$, we take $g = -5$. Then $f = 4 + 2(-5) = 4 - 10 = -6$. The center of the circle is $C(-g, -f) = (5, 6)$. The radius is $r = 6$. The equation of the circle is $(x-5)^2 + (y-6)^2 = 6^2 = 36$. We need to find the square of the length of the chord of the circle on the line $x=1$. The distance from the center $(5,6)$ to the line $x=1$ is $d = |5-1| = 4$. Let the length of the chord be $2L$. Then $L^2 + d^2 = r^2 \Rightarrow L^2 + 4^2 = 6^2 \Rightarrow L^2 + 16 = 36 \Rightarrow L^2 = 20$. The length of the chord is $2L = 2\sqrt{20}$. The square of the length of the chord is $(2L)^2 = (2\sqrt{20})^2 = 4 imes 20 = 80$.

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