The circles x^2+y^2+2ax+c=0 and x^2+y^2+2by+c | vedclass

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(B) For the circle $x^2+y^2+2ax+c=0$,the center is $C_1 = (-a, 0)$ and the radius is $r_1 = \sqrt{a^2-c}$.For the circle $x^2+y^2+2by+c=0$,the center

Vedclass · Accepted Answer

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

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(B) For the circle $x^2+y^2+2ax+c=0$,the center is $C_1 = (-a, 0)$ and the radius is $r_1 = \sqrt{a^2-c}$.For the circle $x^2+y^2+2by+c=0$,the center is $C_2 = (0, -b)$ and the radius is $r_2 = \sqrt{b^2-c}$.Since the circles touch each other externally,the distance between their centers must be equal to the sum of their radii:$d(C_1, C_2) = r_1 + r_2$$\sqrt{(-a-0)^2 + (0-(-b))^2} = \sqrt{a^2-c} + \sqrt{b^2-c}$$\sqrt{a^2+b^2} = \sqrt{a^2-c} + \sqrt{b^2-c}$Squaring both sides:$a^2+b^2 = (a^2-c) + (b^2-c) + 2\sqrt{(a^2-c)(b^2-c)}$$a^2+b^2 = a^2+b^2-2c + 2\sqrt{(a^2-c)(b^2-c)}$$2c = 2\sqrt{(a^2-c)(b^2-c)}$$c = \sqrt{(a^2-c)(b^2-c)}$Squaring again:$c^2 = (a^2-c)(b^2-c)$$c^2 = a^2b^2 - a^2c - b^2c + c^2$$0 = a^2b^2 - c(a^2+b^2)$$c(a^2+b^2) = a^2b^2$Dividing both sides by $a^2b^2c$:$\frac{a^2+b^2}{a^2b^2} = \frac{1}{c}$$\frac{1}{b^2} + \frac{1}{a^2} = \frac{1}{c}$

The circles $x^2+y^2+2ax+c=0$ and $x^2+y^2+2by+c=0$ touch each other externally,if

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