Circle
Common Tangents
IIT-JEE 2002
Grade 11
Question:
If $a > 2b > 0$, the positive value of $m$ for which $y = mx - b\sqrt{1 + m^2}$ is a common tangent to $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$ is:
(1) $\frac{2b}{\sqrt{a^2 - 4b^2}}$
(2) $\frac{\sqrt{a^2 - 4b^2}}{2b}$
(3) $\frac{b}{\sqrt{a^2 - 4b^2}}$
(4) $\frac{b}{a - 2b}$
Step-by-Step Solution
Key Concept: The line is already tangent to $x^2 + y^2 = b^2$. For tangency to $(x - a)^2 + y^2 = b^2$: distance from $(a, 0)$ to $mx - y - b\sqrt{1 + m^2} = 0$ equals $b$. This gives $ma = 2b\sqrt{1 + m^2}$; solve for $m$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)