NTA Abhyas JEE Main2020MathematicsCirclePractice
Let the vertices of a triangle are A = - 3 + 2 sin θ , 4 + 2 cos θ , B = - 3 + 2 cos θ , 4 - 2 sin θ and C = - 3 - 2 sin θ , 4 - 2 cos θ , then the distance between the centroid and the circumcentre of Δ A B C is
Options
- A2 3 units
- B3 2 units
- C1 2 units
- D1 3 units
Correct answer
A. 2 3 units
Step-by-step solution
A , B , C lie on the circle x + 3 2 + y - 4 2 = 2 2 ⇒ circumcentre of Δ A B C is - 3,4 = S & the centroid of Δ A B C is - 3 + 2 3 cos θ , 4 - 2 3 sin θ = G Distance between S & G is 2 3 2 c o s 2 θ + 2 3 2 s i n 2 θ = 2 3 units