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

Find the equation of a circle with center $(4, 3)$ that touches the circle $x^2 + y^2 = 1$ internally.

If the circles given by $S \equiv x^2+y^2-14x+6y+33=0$ and $S^{\prime} \equiv x^2+y^2-a^2=0$ $(a \in N)$ have $4$ common tangents,then the possible number of circles $S^{\prime}=0$ is

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 - 8x + y - 15 = 0$ and $x^2 + y^2 - 4x + 4y - 42 = 0$.

If the circles of same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally,then $a =$

The equation of the direct common tangent of the circles $x^2+y^2-6x-4y-23=0$ and $x^2+y^2+2x+2y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module