If the circles x^2 + y^2 = 4 and x^2 + y^2 - | vedclass

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

Question

(C) The given circles are $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$.For the first circle,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The given circles are $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$.For the first circle,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.For the second circle,the center $C_2 = (5, 0)$ and radius $r_2 = \sqrt{5^2 - \lambda} = \sqrt{25 - \lambda}$.Since the circles touch externally,the distance between their centers must be equal to the sum of their radii: $C_1C_2 = r_1 + r_2$.The distance $C_1C_2 = \sqrt{(5-0)^2 + (0-0)^2} = 5$.So,$5 = 2 + \sqrt{25 - \lambda}$.$3 = \sqrt{25 - \lambda}$.Squaring both sides,$9 = 25 - \lambda$.Therefore,$\lambda = 25 - 9 = 16$.

If the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$ touch externally,then $\lambda$ is equal to:

Similar Questions

The equation of the circle which passes through the point $(3,2)$,bisects the circumference of the circle $x^2+y^2=15$,and cuts the circle $x^2+y^2+4x+6y+3=0$ orthogonally is

If the equation of the common tangent of the circles $x^2+y^2-4x+6y+4=0$ and $x^2+y^2+2x-2y-2=0$ at their point of contact is $ax+by+c=0$,then $\frac{a}{c}=$

The distance between the centres of similitude of the circles $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$ is

If $x^2+y^2-6x-8y+12=0$ and $x^2+y^2-4x+6y+k=0$ cut orthogonally,then $k=$

If the angle between the circles $x^2+y^2-4x-6y-3=0$ and $x^2+y^2+8x-4y+\lambda=0$ is $60^{\circ}$,then a value of $\lambda$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module