Question

If the equation of a circle passing through the point (2, 1) and the points of intersection of the circles $x^{2}+y^{2}+4x-6y-3=0$ and $x^{2}+y^{2}-2x+2y-2=0$ is $x^{2}+y^{2}+2gx+2fy+c=0$, then $2g+f=$

• A. -2c
• B. 2c
• C. c
• D. -c

Hint:

Use $S_1 + \lambda S_2 = 0$ directly for problems involving a point and the intersection of two circles to avoid calculating the common chord first.

Answer 1

The Correct Option is C

Step 1: Use the family of circles.
Any circle through the two intersection points of $S_1=0$ and $S_2=0$ can be written as $S_1+\lambda S_2=0$ for some number $\lambda$.
Step 2: Name the two circles.
Let $S_1=x^2+y^2+4x-6y-3$ and $S_2=x^2+y^2-2x+2y-2$.
Step 3: Use the extra point $(2,1)$.
Put $(2,1)$ into $S_1+\lambda S_2=0$. We get $S_1(2,1)=4$ and $S_2(2,1)=1$, so $4+\lambda=0$, giving $\lambda=-4$.
Step 4: Form the required circle.
$S_1-4S_2=0$ gives $-3x^2-3y^2+12x-14y+5=0$. Divide by $-3$: \[ x^2+y^2-4x+\tfrac{14}{3}y-\tfrac{5}{3}=0. \]
Step 5: Read off $g,f,c$.
Here $2g=-4$, $f=\frac{7}{3}$, and $c=-\frac{5}{3}$.
Step 6: Compute $2g+f$.
\[ 2g+f=-4+\tfrac{7}{3}=-\tfrac{5}{3}=c. \] So $2g+f=c$. \[ \boxed{2g+f=c} \]