If the circles x^2 + y^2 + 2ax + c = 0 and x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The centers of the circles are $C_1(-a, 0)$ and $C_2(0, -b)$.The radii of the circles are $R_1 = \sqrt{a^2 - c}$ and $R_2 = \sqrt{b^2 - c}$.The di

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The centers of the circles are $C_1(-a, 0)$ and $C_2(0, -b)$.The radii of the circles are $R_1 = \sqrt{a^2 - c}$ and $R_2 = \sqrt{b^2 - c}$.The distance between the centers is $C_1C_2 = \sqrt{(-a - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2}$.Since the circles touch each other,the distance between their centers must be equal to the sum of their radii: $R_1 + R_2 = C_1C_2$.$\sqrt{a^2 - c} + \sqrt{b^2 - c} = \sqrt{a^2 + b^2}$.Squaring both sides: $(a^2 - c) + (b^2 - c) + 2\sqrt{(a^2 - c)(b^2 - c)} = a^2 + b^2$.$a^2 + b^2 - 2c + 2\sqrt{(a^2 - c)(b^2 - c)} = a^2 + b^2$.$2\sqrt{(a^2 - c)(b^2 - c)} = 2c$.$\sqrt{(a^2 - c)(b^2 - c)} = c$.Squaring again: $(a^2 - c)(b^2 - c) = c^2$.$a^2b^2 - a^2c - b^2c + c^2 = c^2$.$a^2b^2 = a^2c + b^2c$.Dividing both sides by $a^2b^2c$: $\frac{a^2b^2}{a^2b^2c} = \frac{a^2c}{a^2b^2c} + \frac{b^2c}{a^2b^2c}$.$\frac{1}{c} = \frac{1}{b^2} + \frac{1}{a^2}$.

If the circles $x^2 + y^2 + 2ax + c = 0$ and $x^2 + y^2 + 2by + c = 0$ touch each other,then:

Similar Questions

The coordinates of the radical centre of the three circles $x^2 + y^2 - 4x - 2y + 6 = 0$,$x^2 + y^2 - 2x - 4y - 1 = 0$,and $x^2 + y^2 - 12x + 2y + 30 = 0$ are

The condition for the circles $x^2+y^2+ax+4=0$ and $x^2+y^2+by+4=0$ to touch each other is

If $(-1, -1)$ is the radical centre of the circles $x^2 + y^2 + 2gx - 4y + 4 = 0$,$x^2 + y^2 + 6x + 2fy + 12 = 0$,and $x^2 + y^2 + 10y + 20 = 0$,then $g - f = $

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

The radical centre of the circles $x^2 + y^2 + 4x + 6y = 19$,$x^2 + y^2 = 9$,and $x^2 + y^2 - 2x - 2y = 5$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module