Question

The equation of the circle which passes through the origin and cuts orthogonally each of the circles \( x^{2} + y^{2} - 6x + 8 = 0 \) and \( x^{2} + y^{2} - 2x - 2y = 7 \) is

• A. $3x^{2}+3y^{2}-8x-13y=0$
• B. $3x^{2}+3y^{2}-8x+29y=0$
• C. $3x^{2}+3y^{2}+8x+29y=0$
• D. $3x^{2}+3y^{2}-8x-29y=0$

Hint:

Orthogonality condition: $2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$.

Answer 1

The Correct Option is B

Step 1: General Equation
Let the equation be $x^{2}+y^{2}+2gx+2fy=0$ (passing through origin, so $c=0$).
Step 2: Orthogonality Conditions
Condition: $2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$. With $x^{2}+y^{2}-6x+8=0$: $2g(-3) + 2f(0) = 0 + 8 \Rightarrow -6g = 8 \Rightarrow 2g = -8/3$. With $x^{2}+y^{2}-2x-2y-7=0$: $2g(-1) + 2f(-1) = 0 - 7 \Rightarrow -2g - 2f = -7$.
Step 3: Solve for Coefficients
Substitute $2g = -8/3$: $-(-8/3) - 2f = -7 \Rightarrow 8/3 + 7 = 2f \Rightarrow 29/3 = 2f$.
Step 4: Final Equation
$x^{2}+y^{2} - (8/3)x + (29/3)y = 0 \Rightarrow 3x^{2}+3y^{2}-8x+29y=0$.
Final Answer: (b)