Question

The equation of the circle passing through the point \((2a, 0)\) and whose radical axis is \(x = \frac{a}{2}\) with respect to the circle \(x^2 + y^2 = a^2\), will be

• A. \(x^2 + y^2 - 2ax = 0\)
• B. \(x^2 + y^2 + 2ax = 0\)
• C. \(x^2 + y^2 + 2ay = 0\)
• D. \(x^2 + y^2 - 2ay = 0\)

Hint:

Radical axis of two circles is obtained by subtracting their equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let the required circle be \(x^2 + y^2 + 2gx + 2fy + c = 0\). Radical axis with given circle is obtained by subtracting the equations.

Step 2: Detailed Explanation:
Given circle: \(S_1: x^2 + y^2 = a^2\) or \(x^2 + y^2 - a^2 = 0\). Let required circle: \(S: x^2 + y^2 + 2gx + 2fy + c = 0\). Radical axis: \(S - S_1 = 0\) gives \(2gx + 2fy + c + a^2 = 0\). This is given as \(x = a/2\). So \(f = 0\), and \(2gx + c + a^2 = 0\) must be equivalent to \(x - a/2 = 0\). Thus \(2g = 1\) and \(c + a^2 = -a/2\)? Wait: Comparing coefficients: \(2gx + (c + a^2) = 0\) should be proportional to \(x - a/2 = 0\). So \(2g = k\), \(c + a^2 = -k(a/2)\). Also circle passes through \((2a,0)\): \(4a^2 + 4ag + c = 0\). Choose \(k=2\) for simplicity: Then \(2g = 2 \Rightarrow g=1\), and \(c + a^2 = -2(a/2) = -a \Rightarrow c = -a - a^2\). Substitute in circle equation: \(4a^2 + 4a(1) + (-a - a^2) = 0 \Rightarrow 4a^2 + 4a - a - a^2 = 0 \Rightarrow 3a^2 + 3a = 0 \Rightarrow a(3a+3)=0\) which holds for \(a=0\) or \(a=-1\), not general. So better approach: From radical axis: \(2gx + 2fy + c + a^2 = 0\) and given \(x = a/2\), we have \(f=0\) and \(2gx + c + a^2 = 0\) must be \(x - a/2 = 0\) multiplied by constant. So \(2g = \lambda\), \(c + a^2 = -\lambda a/2\). So \(g = \lambda/2\), \(c = -\lambda a/2 - a^2\). Point \((2a,0)\): \(4a^2 + 2g(2a) + c = 0 \Rightarrow 4a^2 + 4a g + c = 0\). Substitute: \(4a^2 + 4a(\lambda/2) + (-\lambda a/2 - a^2) = 0 \Rightarrow 4a^2 + 2a\lambda - \lambda a/2 - a^2 = 0 \Rightarrow 3a^2 + \frac{3a\lambda}{2} = 0 \Rightarrow 3a(a + \lambda/2) = 0\). For non-zero a, \(\lambda = -2a\). Then \(g = \lambda/2 = -a\), \(c = -\lambda a/2 - a^2 = -(-2a)a/2 - a^2 = a^2 - a^2 = 0\). So circle: \(x^2 + y^2 - 2ax + 0 \cdot y + 0 = 0\) i.e., \(x^2 + y^2 - 2ax = 0\).

Step 3: Final Answer:
\(x^2 + y^2 - 2ax = 0\), which corresponds to option (A).