Question

A circle C passing through the point (1,1) bisects the circumference of the circle \( x^2+y^2-2x=0 \). If C is orthogonal to the circle \( x^2+y^2+2y-3=0 \) then the centre of the circle C is

• A. \( \left(-\frac{5}{2}, 0\right) \)
• B. \( \left(\frac{5}{2}, 0\right) \)
• C. \( \left(0, \frac{5}{2}\right) \)
• D. \( \left(0, -\frac{1}{2}\right) \)

Hint:

For orthogonal circles \( S=0, S'=0 \), the condition is \( 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We set up a system of linear equations for the parameters \( g, f, c \) of the required circle based on the three geometric conditions provided.
Step 2: Detailed Explanation:

Let Circle C: \( x^2+y^2+2gx+2fy+c=0 \). 1. Passes through \( (1,1) \): \( 1 + 1 + 2g + 2f + c = 0 \implies 2g + 2f + c = -2 \). (Eq 1) 2. Bisects circumference of \( S': x^2+y^2-2x=0 \): Common chord \( S - S' = 0 \) passes through center of \( S' \). Center of \( S' \) is \( (1, 0) \). Eq of chord: \( (2g+2)x + 2fy + c = 0 \). Substitute \( (1,0) \): \( (2g+2)(1) + 0 + c = 0 \implies 2g + c = -2 \). (Eq 2) 3. Orthogonal to \( S'': x^2+y^2+2y-3=0 \): Condition: \( 2g(0) + 2f(1) = c + (-3) \implies 2f = c - 3 \implies c = 2f + 3 \). (Eq 3) Solving: Substitute Eq 3 into Eq 2: \( 2g + (2f + 3) = -2 \implies 2g + 2f = -5 \). Substitute Eq 3 into Eq 1: \( 2g + 2f + (2f + 3) = -2 \implies 2g + 4f = -5 \). Subtract: \( (2g+4f) - (2g+2f) = -5 - (-5) \implies 2f = 0 \implies f = 0 \). From \( 2g + 0 = -5 \implies g = -5/2 \). Center \( (-g, -f) = (5/2, 0) \).
Step 3: Final Answer:

The center is \( \left(\frac{5}{2}, 0\right) \).