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

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

Question

(B) The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0, -2)$ and their radii are $r_1 = \sqrt{\lambda^2-2}$ and $r_2 = \sqrt{2}$.The two

Vedclass · Accepted Answer

$\pm 2$

Vedclass · Answer

(B) The centres of the two circles are $C_1(-\lambda, 0)$ and $C_2(0, -2)$ and their radii are $r_1 = \sqrt{\lambda^2-2}$ and $r_2 = \sqrt{2}$.The two circles touch each other if the distance between their centres $C_1C_2$ is equal to the sum or difference of their radii: $C_1C_2 = |r_1 \pm r_2|$.Calculating the distance $C_1C_2 = \sqrt{(-\lambda - 0)^2 + (0 - (-2))^2} = \sqrt{\lambda^2 + 4}$.Setting $C_1C_2 = r_1 + r_2$:$\sqrt{\lambda^2 + 4} = \sqrt{\lambda^2 - 2} + \sqrt{2}$Squaring both sides:$\lambda^2 + 4 = (\lambda^2 - 2) + 2 + 2\sqrt{2(\lambda^2 - 2)}$$\lambda^2 + 4 = \lambda^2 + 2\sqrt{2(\lambda^2 - 2)}$$4 = 2\sqrt{2(\lambda^2 - 2)}$$2 = \sqrt{2(\lambda^2 - 2)}$Squaring again:$4 = 2(\lambda^2 - 2)$$2 = \lambda^2 - 2$$\lambda^2 = 4 \Rightarrow \lambda = \pm 2$.

If the circles $x^2+y^2+2 \lambda x+2=0$ and $x^2+y^2+4y+2=0$ touch each other,then $\lambda=$

Similar Questions

If the circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ touch each other internally,then $\alpha$ is equal to

$A$ circle passes through $(0, 0)$ and $(1, 0)$ and touches the circle ${x^2} + {y^2} = 9$. Find the centre of the circle.

$A$ circle $S$ passes through the points of intersection of the circles $x^2+y^2-2x-3=0$ and $x^2+y^2-2y=0$. If $x+y+1=0$ is a tangent to the circle $S$,then the equation of $S$ is

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

The value of $k$ such that the circles $x^2 + y^2 + kx + 4y + 2 = 0$ and $2(x^2 + y^2) - 4x - 3y + k = 0$ cut orthogonally is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module