Question

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

• A. \( 2x^2 + 2y^2 - 5x + 10y = 0 \)
• B. \( 2x^2 + 2y^2 - 10x + 5y = 0 \)
• C. \( x^2 + y^2 - 2x + 5y = 0 \)
• D. \( x^2 + y^2 - 5x + 2y = 0 \)

Hint:

To find a circle that cuts given circles orthogonally, use the orthogonality condition \( 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2 \). Solve for \( g, f \) and substitute into the general circle equation.

Answer 1

The Correct Option is D

Step 1: General equation of a circle 
The standard equation of a circle is: \[ x^2 + y^2 + 2gx + 2fy + c = 0. \] Since the required circle passes through the origin, substituting \( (0,0) \): \[ 0 + 0 + 2g(0) + 2f(0) + c = 0 \Rightarrow c = 0. \] Thus, the equation simplifies to: \[ x^2 + y^2 + 2gx + 2fy = 0. \] 

Step 2: Condition for orthogonality 
Two circles cut each other orthogonally if the condition: \[ 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2 \] is satisfied, where \( (g_1, f_1) \) and \( (g_2, f_2) \) are the centres of the two given circles. For the first circle: \[ x^2 + y^2 + 6x - 15 = 0. \] Comparing with \( x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0 \): \[ 2g_1 = 6 \Rightarrow g_1 = 3, \quad f_1 = 0, \quad c_1 = -15. \] For the second circle: \[ x^2 + y^2 - 8y - 10 = 0. \] Comparing: \[ g_2 = 0, \quad 2f_2 = -8 \Rightarrow f_2 = -4, \quad c_2 = -10. \]

Step 3: Apply orthogonality condition 
\[ 2g(3) + 2f(0) = -15 + 0. \] \[ 6g = -15 \Rightarrow g = -\frac{5}{2}. \] \[ 2g(0) + 2f(-4) = 0 - 10. \] \[ -8f = -10 \Rightarrow f = \frac{5}{4}. \] 

Step 4: Find the required equation 
Substituting \( g = -\frac{5}{2} \) and \( f = \frac{5}{4} \) into: \[ x^2 + y^2 + 2gx + 2fy = 0. \] \[ x^2 + y^2 - 5x + 2y = 0. \] 

Step 5: Conclusion 
Thus, the correct answer is: \[ \mathbf{x^2 + y^2 - 5x + 2y = 0.} \]