The points $(1, 1), (- 6, 0), (- 2, 2), (- 2, - 8)$ are

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

none

collinear

vertices of a square

concyclic

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let $A (1, 1) B (- 6, 0) C (- 2, 2) and D (- 2, - 8)$

Equation of the circle with $\bar{A B}$ as a diameter is $x^{2} + y^{2} - (1 - 6) x - (1 + 0) y + 1 (- 6) + 1 (0) = 0$

$\Rightarrow x^{2} + y^{2} + 5 x - y - 6 = 0$ . Equation to $A B is x - 7 y + 6 = 0$

Equation of the circle passing through $A, B is x^{2} + y^{2} + 5 x - y - 6 + λ (x - 7 y + 6) = 0$

This passes through $C (- 2, 2) \Rightarrow 4 + 4 + 5 (- 2) - 2 - 6 + λ (- 2 - 7 (2) + 6) = 0$ $\Rightarrow - 10 - 10 λ = 0 \Rightarrow λ = - 1$

Equation of the circle passing through $A, B & C is x^{2} + y^{2} + 4 x + 6 y - 12 = 0 \to (1)$

$(- 2)^{2} + (- 8)^{2} + 4 (- 2) + 6 (- 8) - 12 = 4 + 64 - 8 - 48 - 12 = 0 \Rightarrow D$ lies on the circle (l) Given points are concyclic.

JEE

NEET

Foundation JEE

Foundation NEET

CBSE