JEE PYQ: Circle Question 37
Question 37 - 2019 (10 Jan Shift 2)
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:
(1) $13$
(2) $20$
(3) $-25$
(4) $25$
Show Answer
Answer: (4)
Solution
Center $(-5, -6)$. Area of equilateral triangle $= \frac{3\sqrt{3}}{4}r^2 \cdot ($ side $= r\sqrt{3})$. Actually $\frac{3\sqrt{3}}{4}(\sqrt{3}r)^2 = \frac{3\sqrt{3}}{4} \cdot 3r^2 = 27\sqrt{3}$. So $r^2 = 36$, $r = 6$. $25 + 36 - c = 36 \Rightarrow c = 25$.
BETA
