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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Circle

If the area of an equilateral triangle inscribed in the circle x ^...

If the area of an equilateral triangle inscribed in the circle x2+y2+10x+12y+c=0{x}^{2} + {y}^{2} + 10x + 12y + c = 0 is 273sq.units,27\sqrt{3}sq . units, then cc is equal to:If the area of an equilateral triangle inscribed in the circle x ^ 2 + y ^ 2 + 10x + 12y + c = 0 is 27 3 sq . units,

Options

  1. A2525
  2. B1313
  3. C−25- 25
  4. D2020

Correct answer

A. 2525

Step-by-step solution

If the area of an equilateral triangle inscribed in the circle x ^ 2 + y ^ 2 + 10x + 12y + c = 0 is 27 3 sq . units,

Given, area of the equilateral triangle inscribed in the circle is 27327\sqrt{3}, hence

34a2=273\dfrac{\sqrt{3}}{4}{a}^{2} = 27\sqrt{3} ⇒a=63...(1)⇒ a = 6\sqrt{3} . . . \left(1\right)

Again, rcos30°=a2rcos30 ° = \dfrac{a}{2}

⇒r=6⇒ r = 6

Hence, the radius of a circle, 52+62−c=6\sqrt{{5}^{2} + {6}^{2} - c} = 6

⇒25+36−c=36⇒ 25 + 36 - c = 36

⇒c=25⇒ c = 25

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