Question

If a tangent to the circle $x^2+y^2+2x+2y+1=0$ is radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$, then

• A. $g=\frac{3}{7}$ or $f=4$
• B. $g=\frac{3}{2}$ or $f=\frac{3}{2}$
• C. $g=\frac{3}{5}$ or $f=1$
• D. $g=\frac{3}{4}$ or $f=2$

Hint:

The radical axis of two circles is the locus of points with equal power with respect to both circles. Its equation is $S_1 - S_2 = 0$ (after normalizing coefficients). If a line $Ax+By+C=0$ is tangent to a circle, the distance from the center to the line equals the radius.

Answer 1

The Correct Option is D

Step 1: Find the equation of the radical axis.
Let $S_A \equiv x^2+y^2+2gx+2fy+c=0$. Let $S_B \equiv 2x^2+2y^2+3x+8y+2c=0$. To find the radical axis, we first normalize $S_B$ by dividing by 2: $S_B' \equiv x^2+y^2+\frac{3}{2}x+4y+c=0$. The radical axis is the line $L \equiv S_A - S_B' = 0$. \[ (x^2+y^2+2gx+2fy+c) - (x^2+y^2+\frac{3}{2}x+4y+c) = 0. \] \[ \left(2g-\frac{3}{2}\right)x + (2f-4)y = 0. \]

Step 2: Find the center and radius of the first circle.
Let the first circle be $S_1 \equiv x^2+y^2+2x+2y+1=0$. We can rewrite this by completing the square: $(x^2+2x+1) + (y^2+2y+1) - 1 = 0$. \[ (x+1)^2 + (y+1)^2 = 1. \] The center is $C = (-1,-1)$ and the radius is $R=1$.

Step 3: Apply the tangency condition.
The radical axis $L$ must be tangent to the circle $S_1$. The condition for tangency is that the perpendicular distance from the center of the circle to the line is equal to the radius. Distance from $C(-1,-1)$ to line $\left(2g-\frac{3}{2}\right)x + (2f-4)y = 0$ must be 1. \[ \frac{\left|\left(2g-\frac{3}{2}\right)(-1) + (2f-4)(-1)\right|}{\sqrt{\left(2g-\frac{3}{2}\right)^2 + (2f-4)^2}} = 1. \] \[ \left|-2g+\frac{3}{2} - 2f+4\right| = \sqrt{\left(2g-\frac{3}{2}\right)^2 + (2f-4)^2}. \]

Step 4: Observe a property of the radical axis and simplify.
The equation of the radical axis, $\left(2g-\frac{3}{2}\right)x + (2f-4)y = 0$, has no constant term. This means it always passes through the origin $(0,0)$. So, the problem is equivalent to finding the tangents from the origin to the circle $(x+1)^2+(y+1)^2=1$. By inspection, the circle passes through $(-1,0)$ and $(0,-1)$ and has center $(-1,-1)$. The lines $y=0$ (x-axis) and $x=0$ (y-axis) are tangent to this circle at $(-1,0)$ and $(0,-1)$ respectively. The radical axis must be one of these two lines. Case 1: The radical axis is the line $x=0$. This occurs if the coefficient of $y$ is non-zero and the coefficient of $x$ is zero. \[ 2g-\frac{3}{2} = 0 \implies 2g = \frac{3}{2} \implies g=\frac{3}{4}. \] Case 2: The radical axis is the line $y=0$. This occurs if the coefficient of $x$ is non-zero and the coefficient of $y$ is zero. \[ 2f-4 = 0 \implies 2f = 4 \implies f=2. \] Therefore, the condition is either $g=3/4$ or $f=2$.