Question

Find the area of an equilateral triangle inscribed in the circle x^2+y^2- 6x + 2y – 15 = 0.

Answer 1

Given equation of circle is: 

x2 + y2 - 6x + 2y - 15 = 0

Centre O = (3, -1)

radius r = √{32 + (-1)2 -(-15)} = √{9 + 1 + 15 } = √25 = 5 

Let ABC is an equilateral triangle inscribed in the give circle, then its altitude AD passes through the cneter of the circle

and D is the mid-point of BC

Also ∠OBD = 60

From right angle triangle OBD,

      sin 60 = BD/OB

=> √3/2 = BD/r

=> BD = r*√3/2

Also, BC = 2*BD = 2*r*√3/2 = √3r

Now, area of the triangle ABC = √3/4 * a2    {a is the side of the triangle ABC} 

                                          = √3/4 * (√3r)2

                                          = 3√3/4 * r2

                                          = 3√3/4 * 52

                                          = 3√3/4 * 25

                                          = 75√3/4 square unit

Answer 2

Step-by-step explanation:

$hey \: hope \: it \: helps \: you$

Given equation of circle is: 

x2 + y2 - 6x + 2y - 15 = 0

Centre O = (3, -1)

radius r = √{32 + (-1)2 -(-15)} = √{9 + 1 + 15 } = √25 = 5 

Let ABC is an equilateral triangle inscribed in the give circle, then its altitude AD passes through the canter of the circle

and D is the mid-point of BC

Also ∠OBD = 60

From right angle triangle OBD,

     sin 60 = BD/OB

=> √3/2 = BD/r

=> BD = r*√3/2

Also, BC = 2*BD = 2*r*√3/2 = √3r

Now, area of the triangle ABC = √3/4 * a2    {a is the side of the triangle ABC} 

                                         = √3/4 * (√3r)2

                                         = 3√3/4 * r2

                                         = 3√3/4 * 52

                                         = 3√3/4 * 25

                                       

 = 75√3/4square unit