Question:

The equation of the circle passing through the origin and cutting orthogonally the circles \( x^2 + y^2 + 6x - 15 = 0 \) and \( x^2 + y^2 - 8y - 10 = 0 \) is:

Show Hint

Circle Orthogonality}
Use condition: \( 2gg_1 + 2ff_1 = c + c_1 \)
Ensure the required circle satisfies passing through origin: no constant term
Scale up to match option formats
Updated On: May 19, 2025
  • \( 2(x^2 + y^2) - 10x + 5y = 0 \)
  • \( 2(x^2 - y^2) + 10x - 5y = 0 \)
  • \( 2(x^2 - y^2) + 10x + 5y = 0 \)
  • \( 2(x^2 - y^2) - 10x - 5y = 0 \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Let the required circle be: \[ x^2 + y^2 + 2gx + 2fy = 0 \quad \text{(passes through origin)} \] Circle 1: \( x^2 + y^2 + 6x - 15 = 0 \) Orthogonal circles satisfy: \[ 2gg_1 + 2ff_1 = c + c_1 \] So for circle 1: \[ g_1 = 3,\ f_1 = 0,\ c_1 = -15,\ c = 0 \Rightarrow 6g + 0 = -15 \Rightarrow g = -\frac{5}{2} \] Circle 2: \( x^2 + y^2 - 8y - 10 = 0 \Rightarrow g_1 = 0,\ f_1 = -4,\ c_1 = -10 \) \[ 0 + (-8f) = -10 \Rightarrow f = \frac{5}{4} \] Now plug into: \[ x^2 + y^2 -5x + \frac{5}{2}y = 0 \Rightarrow \text{Multiply by 2: } 2x^2 + 2y^2 -10x + 5y = 0 \]
Was this answer helpful?
0
0