AP EAMCET · Maths · Circle

The number of circles that touch all the three lines \(x+y-1=0, x-y-1=0\) and \(y+1=0\) is

A \(2\)
B \(3\)
C \(4\)
D \(1\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

Given lines are \[ x+y-1=0, x-y-1=0 \text { and } y+1=0 \] These lines form a triangle the total number of circles that touches all the three sides of a triangle are 4. \([\because\) One is incircle of a triangle and when we extend the sides of a triangle we will get the three…

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.

Same subject

Which one of the following is false?

AP EAMCET 2024 Easy

If \(\cos \alpha+\cos \beta=a\) and \(\sin \alpha+\sin \beta=b\), then match the items given in List \(-A\) with those of their values in List-B

List-A	Liat-B
(I) \(\tan \left(\frac{\alpha+\beta}{2}\right)=\)	(a) \(\frac{ b }{ a }\)
(II) \(\cos (\alpha+\beta)=\)	(b) \(\frac{2 a b}{a^2+b^2}\)
(III) \(\sin (\alpha+\beta)^{-}\)	(c) \(\frac{2 a b}{a^2-b^2}\)
(IV) \(\tan (\alpha+\beta)=\)	(d) \(\frac{a^2-b^2}{a^2+b^2}\)
	(e) \(\frac{a^2+b^2}{a^2-b^2}\)

AP EAMCET 2023 Medium

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