AP EAMCET · Maths · Circle

If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4 y+2=0$ touch each other, then $\lambda=$

A $\pm 1$
B $\pm 2$
C $\pm 3$
D $\pm 4$

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Answer & Solution

Correct Answer

(B) $\pm 2$

Step-by-step Solution

Detailed explanation

The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0,-2)$ and their radii are $\sqrt{\lambda^2-2}$ and $\sqrt{2}$. So, the two circles will touch each other, if…

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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Match the following.

List - $I$ Reaction		List - $II$ Catalyst
$(A)$	$N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)$	$(I)$	$Ni$
$(B)$	$2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l)$	$(II)$	$Pt$
$(C)$	$CO (g) + 3 H_{2} (g) ⟶ {CH}_{4} (g) + H_{2} O (g)$	$(III)$	$ZnO - {Cr}_{2} O_{3}$
$(D)$	$CO (g) + 2 H_{2} (g) ⟶ {CH}_{3} OH (g)$	$(IV)$	$Fe$

The correct answer is:

AP EAMCET 2019 Easy

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If the circles \(x^2+y^2+2 \lambda x+2=0\) and \(x^2+y^2+4 y+2=0\) touch each other, then \(\lambda=\)

Answer & Solution

Step-by-step Solution

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