AP EAMCET · Maths · Circle

Given that $a > 2 b > 0$ and that the line $y = m x - b \sqrt{1 + m^{2}}$ is a common tangent to the circles $x^{2} + y^{2} = b^{2}$ and $(x - a)^{2} + y^{2} = b^{2}$ . Then the positive value of $m$ is

A $\frac{2 b}{a - 2 b}$
B $\frac{b}{a - 2 b}$
C $\frac{\sqrt{a^{2} - 4 b^{2}}}{2 b}$
D $\frac{2 b}{\sqrt{a^{2} - 4 b^{2}}}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{2 b}{\sqrt{a^{2} - 4 b^{2}}}$

Step-by-step Solution

Detailed explanation

The equation of circles are given as, x2+y2=b2 (x-a)2+y2=b2 The equation of the comment tangent for 1st is, y=mx-b1+m2 The equation of the common tangent for 2nd circle is, y=mx-a+b1+m2 Equate the common tangents, mx-b1+m2=mx-am+b1+m2 2b1+m2=am a2m2=4b2+4b2m2 m2a2-4b2=4b2 The…

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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Given that a>2b>0 and that the line y=mx-b1+m2 is a common tangent to the circles x2+y2=b2 and (x-a)2+y2=b2. Then the positive value of m is

Step-by-step Solution

Given that $a > 2 b > 0$ and that the line $y = m x - b \sqrt{1 + m^{2}}$ is a common tangent to the circles $x^{2} + y^{2} = b^{2}$ and $(x - a)^{2} + y^{2} = b^{2}$ . Then the positive value of $m$ is