TS EAMCET · Maths · Circle

If the chord $L \equiv y - m x - 1 = 0$ of the circle $S \equiv x^{2} + y^{2} - 1 = 0$ touches the circle $S_{1} = x^{2} + y^{2} - 4 x + 1 = 0$ , then the possible points for which $L = 0$ is a chord of contact of $S = 0$ are

A $(2 \pm \sqrt{6}, 0)$
B $(2 \pm \sqrt{6}, 1)$
C $(2, 2)$
D $(\sqrt{6}, 1)$

Verified Solution

Answer & Solution

Correct Answer

(B) $(2 \pm \sqrt{6}, 1)$

Step-by-step Solution

Detailed explanation

L≡y-mx-1=0 of circle S≡x2+y2-1=0 touches the circle S1≡x2+y2-4x+1. Consider line, L≡-mx+y-1=0 and circle S1≡x2+ y2-4x+1. ∴ Centre of S1 is 2,0. If the straight line touches the circle then, 22-1=-m2+0-1-m2+1 ⇒3=-2m-11+m2…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If the chord L≡y-mx-1=0 of the circle S≡x2+y2-1=0 touches the circle S1=x2+y2-4x+1=0, then the possible points for which L=0 is a chord of contact of S=0 are

Step-by-step Solution

If the chord $L \equiv y - m x - 1 = 0$ of the circle $S \equiv x^{2} + y^{2} - 1 = 0$ touches the circle $S_{1} = x^{2} + y^{2} - 4 x + 1 = 0$ , then the possible points for which $L = 0$ is a chord of contact of $S = 0$ are