Question

If the circles $x^2+y^2 - 4x + 2fy + 1 = 0$ and $x^2+y^2 + 2gx - 4y - 1 = 0$ cut orthogonally, then $r_1^2 + r_2^2 - 8=$

• A. $g^2$
• B. $-f^2$
• C. $2g^2$
• D. $-2f^2$

Hint:

Orthogonal Circles.
Use: $2g_1g_2 + 2h_1h_2 = c_1 + c_2$ to get relation, and plug back into expression as needed.

Answer 1

The Correct Option is C

Radius squared of first circle: $r_1^2 = g_1^2 + h_1^2 - c_1 = 4 + f^2 - 1 = f^2 + 3$
Radius squared of second circle: $r_2^2 = g^2 + 4 + 1 = g^2 + 5$ Using orthogonality condition: \[ 2(g_1g_2 + h_1h_2) = c_1 + c_2 \Rightarrow 2(-2g + f(-2)) = 0 \Rightarrow g + f = 0 \Rightarrow f = -g \] Now, \[ r_1^2 + r_2^2 - 8 = f^2 + g^2 + 3 + 5 - 8 = f^2 + g^2 = 2g^2 \]