Question

The condition that the straight line \(cx - by + b^2 = 0\) may touch the circle \(x^2 + y^2 = ax + by\) is

• A. \(abc = 1\)
• B. \(a = c\)
• C. \(b = ac\)
• D. None of these

Hint:

For tangency of a line to a circle, equate perpendicular distance from center to line with radius.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Condition for tangency: perpendicular distance from center to line equals radius.

Step 2: Detailed Explanation:
Circle: \(x^2 + y^2 - ax - by = 0\). Center = \(\left(\frac{a}{2}, \frac{b}{2}\right)\), radius = \(\sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \frac{\sqrt{a^2 + b^2}}{2}\).
Line: \(cx - by + b^2 = 0\).
Distance from center = \(\frac{|c \cdot \frac{a}{2} - b \cdot \frac{b}{2} + b^2|}{\sqrt{c^2 + b^2}} = \frac{|\frac{ac}{2} - \frac{b^2}{2} + b^2|}{\sqrt{c^2 + b^2}} = \frac{|\frac{ac}{2} + \frac{b^2}{2}|}{\sqrt{c^2 + b^2}} = \frac{|ac + b^2|}{2\sqrt{c^2 + b^2}}\).
Set equal to radius: \(\frac{|ac + b^2|}{2\sqrt{c^2 + b^2}} = \frac{\sqrt{a^2 + b^2}}{2}\)
\(\implies |ac + b^2| = \sqrt{(c^2 + b^2)(a^2 + b^2)}\)
Squaring: \((ac + b^2)^2 = (c^2 + b^2)(a^2 + b^2)\)
\(a^2c^2 + 2ab^2c + b^4 = a^2c^2 + a^2b^2 + b^2c^2 + b^4\)
\(2ab^2c = a^2b^2 + b^2c^2\)
If \(b \neq 0\), divide by \(b^2\): \(2ac = a^2 + c^2 \implies (a - c)^2 = 0 \implies a = c\).
So condition is \(a = c\). Option (B) says \(a = c\).

Step 3: Final Answer:
Option (B) \(a = c\). (Wait, re-evaluating: The derivation shows \(a=c\). Option (B) is \(a=c\). So answer should be (B), not (D). Let me correct: The correct condition is \(a=c\).)