Question

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

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

Hint:

Radical axis: Subtract circle equations. Tangency condition: Perpendicular distance from center = radius.

Answer 1

The Correct Option is C

The radical axis of the given circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ is given by $(x^2+y^2+2gx+2fy+c) - \frac{1}{2}(2x^2+2y^2+3x+8y+2c) = 0$, which simplifies to $4gx+4fy-3x-8y=0$ or $(4g-3)x+(4f-8)y=0$. This line touches the circle $x^2+y^2+2x+2y+1=0$. The condition for a line $lx+my+n=0$ to touch a circle $x^2+y^2+2gx+2fy+c=0$ is $(lg+mf+n)^2 = (l^2+m^2)(g^2+f^2-c)$. In our case, $l = (4g-3)$, $m = (4f-8)$, $n=0$. The circle's center is $(-1, -1)$, and its radius is 1. The perpendicular distance from the center $(-1, -1)$ to the line $(4g-3)x + (4f-8)y = 0$ is given by: $\frac{|(4g-3)(-1)+(4f-8)(-1)|}{\sqrt{(4g-3)^2+(4f-8)^2}} = 1$ Squaring both sides and simplifying, we get: $(4g+4f-11)^2 = (4g-3)^2+(4f-8)^2$ $16(g^2+f^2+2gf) + 121 - 88(g+f) = 16g^2-24g+9+16f^2-64f+64$ $32gf+121-88(g+f)=-24g-64f+73$ $32gf - 64g - 24f + 48 = 0$ $4(8gf - 16g - 6f + 12) = 0$ $8gf - 16g - 6f + 12 = 0$ $8g(f-2)-6(f-2)=0$ $(f-2)(8g-6) = 0$ Therefore, either $f=2$ or $g = \frac{6}{8} = \frac{3}{4}$.