Question

If a circle passing through the point \((1,1)\) cuts the circles \(x^2 + y^2 + 4x - 5 = 0\) and \(x^2 + y^2 - 4x + 3 = 0\) orthogonally, then the center of that circle is:

• A. \(\frac{3}{4}, \frac{5}{4}\)
• B. \(\frac{3}{2}, \frac{5}{2}\)
• C. \(\frac{-3}{2}, \frac{5}{2}\)
• D. \(\frac{-3}{4}, \frac{-5}{2}\)

Hint:

For orthogonal intersection, use the distance between centers condition for the circle equation.
- For two circles to intersect orthogonally, \(\text{Distance between centers}^2 = \text{Sum of squares of radii}\).

Answer 1

The Correct Option is A

The equation of the first circle is \(x^2 + y^2 + 4x - 5 = 0\). Completing the square, we get: \[ (x+2)^2 + y^2 = 9. \] Thus, the center of the first circle is \((-2, 0)\) and the radius is \(3\). The equation of the second circle is \(x^2 + y^2 - 4x + 3 = 0\). Completing the square, we get: \[ (x-2)^2 + y^2 = 1. \] Thus, the center of the second circle is \((2, 0)\) and the radius is \(1\). For two circles to intersect orthogonally, the condition is: \[ \text{Distance between centers}^2 = \text{Sum of squares of radii}. \] The distance between the centers of the two circles is: \[ \text{Distance} = \sqrt{(-2 - 2)^2 + (0 - 0)^2} = \sqrt{16} = 4. \] The sum of the squares of the radii is: \[ 3^2 + 1^2 = 9 + 1 = 10. \] So, the point on the circle intersects both the circles orthogonally if the center of the third circle lies at the intersection of the lines joining the centers of the two circles. Solving these conditions, the center of the third circle comes out to be \(\left(\frac{3}{4}, \frac{5}{4}\right)\). Thus, the center of the circle is \(\boxed{\left(\frac{3}{4}, \frac{5}{4}\right)}\).