Question:
If the area of an equilateral triangle inscribed in the circle \(x^2 + y^2 + 10x + 12y + c = 0\) is \(27\sqrt{3}\) sq units then c is equal to :
If the area of an equilateral triangle inscribed in the circle \(x^2 + y^2 + 10x + 12y + c = 0\) is \(27\sqrt{3}\) sq units then c is equal to :
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Area of an equilateral triangle inscribed in a circle of radius R is \(\frac{3\sqrt3}{4}R^2\); find R, then use \(R^2 = g^2 + f^2 - c\).
Updated On: Nov 5, 2026
- 20
- 25
- \(-25\)
- 13
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The Correct Option is B
Solution and Explanation
For an equilateral triangle with circumradius R, side \(a = \sqrt3 R\) and area \(= \frac{\sqrt3}{4}a^2 = \frac{3\sqrt3}{4}R^2\).
\(\frac{3\sqrt3}{4}R^2 = 27\sqrt3 \Rightarrow R^2 = 36\).
For the circle \(x^2 + y^2 + 10x + 12y + c = 0\): \(g = 5, f = 6\), so \(R^2 = g^2 + f^2 - c = 25 + 36 - c = 61 - c\).
\(61 - c = 36 \Rightarrow c = 25\).
Answer: (2) 25
\(\frac{3\sqrt3}{4}R^2 = 27\sqrt3 \Rightarrow R^2 = 36\).
For the circle \(x^2 + y^2 + 10x + 12y + c = 0\): \(g = 5, f = 6\), so \(R^2 = g^2 + f^2 - c = 25 + 36 - c = 61 - c\).
\(61 - c = 36 \Rightarrow c = 25\).
Answer: (2) 25
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