SATHEE: Circle 1 Question 5

Circle 1 Question 5

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

(2019 Main, 10 Jan II)

(a) 20

(b) -25

(d) 25

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Answer:

Correct Answer: 5. (d)

Solution:

Clearly, centre of the circumscribed circle is the centroid $(G)$ of the equilateral triangle $A B C$ .

$[∵$ in an equilateral triangle circumcentre and centroid coincide]

Also, we know that

$△ A G B ≅ △ B G C ≅ △ C G A$ [by SAS congruence rule]

$∴ ar (△ A B C) = 3 ar (△ A G B)$

$\begin{aligned} = 3 & \frac{1}{2} r^{2} \sin 120^{\circ} \\ [∵ area of triangle = \frac{1}{2} a b \sin (∠ C)] \end{aligned}$

$∵ ar (△ A B C) = 27 \sqrt{3}$

[given]

$∴ \frac{3}{2} r^{2} \frac{\sqrt{3}}{2} = 27 \sqrt{3}$

$[\sin 120^{\circ} = \sin (180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}]$

$\Rightarrow r^{2} = 4 \times 9$

$\Rightarrow r = 6$

Now, radius of circle,

$\begin{aligned} r & = \sqrt{g^{2} + f^{2} - c} \\ \Rightarrow 6 & = \sqrt{25 + 36 - c} \end{aligned}$

[ $∵$ in the given equation of circle $2 g = 10$ and $2 f = 12 \Rightarrow g = 5$ and $f = 6$ ]

$\begin{array}{lc} \Rightarrow & 36 = 25 + 36 - c \\ \Rightarrow & c = 25 \end{array}$

Circle 1 Question 5

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

Answer:

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Circle 1 Question 5

5. If the area of an equilateral triangle inscribed in the circle, x2+y2+10x+12y+c=0 is 273 sq units, then c is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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