When do the circles represented by the equati | vedclass

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

Question

(C) The given equations are $S_1: x^{2} + y^{2} - 2ax + c^{2} = 0$ and $S_2: x^{2} + y^{2} - 2by + c^{2} = 0$.For $S_1$,the center $C_1 = (a, 0)$ and

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given equations are $S_1: x^{2} + y^{2} - 2ax + c^{2} = 0$ and $S_2: x^{2} + y^{2} - 2by + c^{2} = 0$.For $S_1$,the center $C_1 = (a, 0)$ and radius $r_1 = \sqrt{a^{2} - c^{2}}$.For $S_2$,the center $C_2 = (0, b)$ and radius $r_2 = \sqrt{b^{2} - c^{2}}$.Two circles touch each other externally if the distance between their centers is equal to the sum of their radii,i.e.,$C_1C_2 = r_1 + r_2$.$C_1C_2 = \sqrt{(a-0)^{2} + (0-b)^{2}} = \sqrt{a^{2} + b^{2}}$.So,$\sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c^{2}} + \sqrt{b^{2} - c^{2}}$.Squaring both sides: $a^{2} + b^{2} = (a^{2} - c^{2}) + (b^{2} - c^{2}) + 2\sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.$a^{2} + b^{2} = a^{2} + b^{2} - 2c^{2} + 2\sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.$2c^{2} = 2\sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.$c^{2} = \sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.Squaring again: $c^{4} = (a^{2} - c^{2})(b^{2} - c^{2}) = a^{2}b^{2} - a^{2}c^{2} - b^{2}c^{2} + c^{4}$.$a^{2}c^{2} + b^{2}c^{2} = a^{2}b^{2}$.Dividing by $a^{2}b^{2}c^{2}$,we get $\frac{1}{b^{2}} + \frac{1}{a^{2}} = \frac{1}{c^{2}}$.

When do the circles represented by the equations $x^{2} + y^{2} + c^{2} = 2ax$ and $x^{2} + y^{2} + c^{2} = 2by$ touch each other externally?

Similar Questions

$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$

For the circle $x-2=5 \cos \theta, y+1=5 \sin \theta$ where $\theta$ is the parameter,the line $x=1+\frac{r}{2}, y=-2+\frac{\sqrt{3}}{2} r$ where $r$ is the parameter,is a

Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

The power of a point $(2,0)$ with respect to a circle $S$ is $-4$ and the length of the tangent drawn from the point $(1,1)$ to $S$ is $2$. If the circle $S$ passes through the point $(-1,-1)$,then the radius of the circle $S$ is

If the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,then $a$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module