Question:
Let the centre of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ 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 $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is __________.
Let the centre of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ 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 $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is __________.
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Find the radius using the area of the equilateral triangle, determine the centre using the line equation and radius formula, then calculate the distance from the centre to $x=1$ to find the chord length.
Updated On: Apr 9, 2026
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Correct Answer: 80
Solution and Explanation
We start by relating the area of the inscribed equilateral triangle to the radius of the circle. Let $R$ be the circumradius of the triangle (which is the radius of the circle). The area of an equilateral triangle in terms of its circumradius $R$ is given by:
$$ \text{Area} = \frac{3\sqrt{3}}{4} R^2 $$
Given that the area is $27\sqrt{3}$, we have:
$$ 27\sqrt{3} = \frac{3\sqrt{3}}{4} R^2 \Rightarrow R^2 = \frac{27 \times 4}{3} = 36 \Rightarrow R = 6 $$
The general equation of the circle is $x^2 + y^2 + 2gx + 2fy + 25 = 0$. Let the centre be $(h, k) = (-g, -f)$. Since it is in the first quadrant, $h>0$ and $k>0$.
The radius squared is given by $R^2 = h^2 + k^2 - c$. Here, $c = 25$. So:
$$ 36 = h^2 + k^2 - 25 \Rightarrow h^2 + k^2 = 61 $$
We are also told that the centre $(h, k)$ lies on the line $2x - y = 4$, so:
$$ 2h - k = 4 \Rightarrow k = 2h - 4 $$
Substituting $k$ into the equation $h^2 + k^2 = 61$:
$$ h^2 + (2h - 4)^2 = 61 $$
$$ h^2 + 4h^2 - 16h + 16 = 61 \Rightarrow 5h^2 - 16h - 45 = 0 $$
Solving this quadratic equation:
$$ h = \frac{16 \pm \sqrt{(-16)^2 - 4(5)(-45)}}{2(5)} = \frac{16 \pm \sqrt{256 + 900}}{10} = \frac{16 \pm \sqrt{1156}}{10} = \frac{16 \pm 34}{10} $$
Since $h>0$, we take $h = \frac{16+34}{10} = 5$. Then $k = 2(5) - 4 = 6$. The centre is $(5, 6)$ and $R = 6$.
The length of the chord of a circle intercepted by a line is $2\sqrt{R^2 - d^2}$, where $d$ is the perpendicular distance from the centre to the line. The line is $x = 1$ (or $x - 1 = 0$).
$$ d = \frac{|5 - 1|}{\sqrt{1^2 + 0^2}} = 4 $$
The length of the chord $L = 2\sqrt{6^2 - 4^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$.
The square of the length of the chord is $L^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$.
$$ \text{Area} = \frac{3\sqrt{3}}{4} R^2 $$
Given that the area is $27\sqrt{3}$, we have:
$$ 27\sqrt{3} = \frac{3\sqrt{3}}{4} R^2 \Rightarrow R^2 = \frac{27 \times 4}{3} = 36 \Rightarrow R = 6 $$
The general equation of the circle is $x^2 + y^2 + 2gx + 2fy + 25 = 0$. Let the centre be $(h, k) = (-g, -f)$. Since it is in the first quadrant, $h>0$ and $k>0$.
The radius squared is given by $R^2 = h^2 + k^2 - c$. Here, $c = 25$. So:
$$ 36 = h^2 + k^2 - 25 \Rightarrow h^2 + k^2 = 61 $$
We are also told that the centre $(h, k)$ lies on the line $2x - y = 4$, so:
$$ 2h - k = 4 \Rightarrow k = 2h - 4 $$
Substituting $k$ into the equation $h^2 + k^2 = 61$:
$$ h^2 + (2h - 4)^2 = 61 $$
$$ h^2 + 4h^2 - 16h + 16 = 61 \Rightarrow 5h^2 - 16h - 45 = 0 $$
Solving this quadratic equation:
$$ h = \frac{16 \pm \sqrt{(-16)^2 - 4(5)(-45)}}{2(5)} = \frac{16 \pm \sqrt{256 + 900}}{10} = \frac{16 \pm \sqrt{1156}}{10} = \frac{16 \pm 34}{10} $$
Since $h>0$, we take $h = \frac{16+34}{10} = 5$. Then $k = 2(5) - 4 = 6$. The centre is $(5, 6)$ and $R = 6$.
The length of the chord of a circle intercepted by a line is $2\sqrt{R^2 - d^2}$, where $d$ is the perpendicular distance from the centre to the line. The line is $x = 1$ (or $x - 1 = 0$).
$$ d = \frac{|5 - 1|}{\sqrt{1^2 + 0^2}} = 4 $$
The length of the chord $L = 2\sqrt{6^2 - 4^2} = 2\sqrt{36 - 16} = 2\sqrt{20}$.
The square of the length of the chord is $L^2 = (2\sqrt{20})^2 = 4 \times 20 = 80$.
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