Question

The circles whose equations are \(x^2 + y^2 + c^2 = 2ax\) and \(x^2 + y^2 + c^2 - 2by = 0\) will touch each other externally if

• A. \(\frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a^2}\)
• B. \(\frac{1}{c^2} + \frac{1}{a^2} = \frac{1}{b^2}\)
• C. \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\)
• D. None of these

Hint:

Use distance formula and condition for external tangency.

Answer 1

The Correct Option is C

Step 1: Formula / Definition}
\[ C_1(a, 0), r_1 = \sqrt{a^2 - c^2}; \quad C_2(0, b), r_2 = \sqrt{b^2 - c^2} \]
Step 2: Calculation / Simplification}
External touch: \(C_1C_2 = r_1 + r_2\)
\(\sqrt{a^2 + b^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2}\)
Square: \(a^2 + b^2 = a^2 - c^2 + b^2 - c^2 + 2\sqrt{(a^2-c^2)(b^2-c^2)}\)
\(2c^2 = 2\sqrt{a^2b^2 - c^2(a^2+b^2) + c^4}\)
\(c^4 = a^2b^2 - c^2(a^2+b^2) + c^4\)
\(a^2b^2 = c^2(a^2+b^2) \Rightarrow \frac{1}{c^2} = \frac{1}{a^2} + \frac{1}{b^2}\)
Step 3: Final Answer
\[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \]