Question

The two circles x²+y²+2ax+c=0 and x²+y²+2by+c=0 touch if $\frac{1}{a^2}+\frac{1}{b^2}=$

• A. 1/c
• B. c
• C. 1/c²
• D. c²

Answer 1

Answer: (A)

1/c

Answer 2

The equations of the two circles are given by: Circle 1: $x^2 + y^2 + 2ax + c = 0$ Circle 2: $x^2 + y^2 + 2by + c = 0$

The center and radius of a circle with the equation $x^2 + y^2 + 2gx + 2fy + d = 0$ are $(-g, -f)$ and $\sqrt{g^2 + f^2 - d}$ respectively.

For Circle 1: Center $C_1 = (-a, 0)$ Radius $r_1 = \sqrt{a^2 - c}$

For Circle 2: Center $C_2 = (0, -b)$ Radius $r_2 = \sqrt{b^2 - c}$

The distance between the centers $C_1$ and $C_2$ is: $d(C_1, C_2) = \sqrt{(0 - (-a))^2 + (-b - 0)^2} = \sqrt{a^2 + b^2}$

For the two circles to touch, the distance between their centers must be equal to the sum or the absolute difference of their radii: $d(C_1, C_2) = r_1 + r_2$ or $d(C_1, C_2) = |r_1 - r_2|$

Squaring both sides: $d(C_1, C_2)^2 = (r_1 \pm r_2)^2$ $a^2 + b^2 = (\sqrt{a^2 - c} \pm \sqrt{b^2 - c})^2$ $a^2 + b^2 = (a^2 - c) + (b^2 - c) \pm 2\sqrt{(a^2 - c)(b^2 - c)}$ $a^2 + b^2 = a^2 + b^2 - 2c \pm 2\sqrt{(a^2 - c)(b^2 - c)}$ $0 = -2c \pm 2\sqrt{(a^2 - c)(b^2 - c)}$ $2c = \pm 2\sqrt{(a^2 - c)(b^2 - c)}$ $c = \pm \sqrt{(a^2 - c)(b^2 - c)}$

Squaring both sides again: $c^2 = (a^2 - c)(b^2 - c)$ $c^2 = a^2b^2 - a^2c - b^2c + c^2$ $0 = a^2b^2 - a^2c - b^2c$ $a^2c + b^2c = a^2b^2$

Divide by $a^2b^2c$ (assuming $a, b, c \neq 0$): $\frac{a^2c}{a^2b^2c} + \frac{b^2c}{a^2b^2c} = \frac{a^2b^2}{a^2b^2c}$ $\frac{1}{b^2} + \frac{1}{a^2} = \frac{1}{c}$

Therefore, $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}$.