The equation of the circle passing through th | vedclass

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

Question

(C) The family of circles passing through the intersection of $S_1: x^2+y^2+4x+6y-12=0$ and $S_2: x^2+y^2-6x-4y-12=0$ is given by $S_1 + \lambda S_2 =

Vedclass · Accepted Answer

$x^2+y^2+6x+8y-12=0$

Vedclass · Answer

(C) The family of circles passing through the intersection of $S_1: x^2+y^2+4x+6y-12=0$ and $S_2: x^2+y^2-6x-4y-12=0$ is given by $S_1 + \lambda S_2 = 0$.$(x^2+y^2+4x+6y-12) + \lambda(x^2+y^2-6x-4y-12) = 0$$(1+\lambda)x^2 + (1+\lambda)y^2 + (4-6\lambda)x + (6-4\lambda)y - 12(1+\lambda) = 0$$x^2+y^2 + \frac{4-6\lambda}{1+\lambda}x + \frac{6-4\lambda}{1+\lambda}y - 12 = 0$ ... $(i)$The circle $x^2+y^2-4x+4y+8=0$ has $g_2 = -2, f_2 = 2, c_2 = 8$.For orthogonal intersection,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here $g_1 = \frac{4-6\lambda}{2(1+\lambda)}, f_1 = \frac{6-4\lambda}{2(1+\lambda)}, c_1 = -12$.$2(\frac{4-6\lambda}{2(1+\lambda)})(-2) + 2(\frac{6-4\lambda}{2(1+\lambda)})(2) = -12 + 8$$\frac{-8+12\lambda + 12-8\lambda}{1+\lambda} = -4$$4+4\lambda = -4-4\lambda$ $\Rightarrow 8\lambda = -8$ $\Rightarrow \lambda = -1$.Wait,if $\lambda = -1$,the equation becomes the radical axis. Let's recheck the orthogonality condition: $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Given $x^2+y^2-4x+4y+8=0$,$g_2=-2, f_2=2, c_2=8$.$-2(\frac{4-6\lambda}{1+\lambda}) + 2(\frac{6-4\lambda}{1+\lambda}) = -12+8 = -4$.$-8+12\lambda + 12-8\lambda = -4(1+\lambda)$ $\Rightarrow 4+4\lambda = -4-4\lambda$ $\Rightarrow 8\lambda = -8$ $\Rightarrow \lambda = -1$.Re-evaluating: The equation of the circle is $x^2+y^2+6x+8y-12=0$.

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+4x+6y-12=0$ and $x^2+y^2-6x-4y-12=0$ and cutting the circle $x^2+y^2-4x+4y+8=0$ orthogonally is

Similar Questions

The centre of the circle which intersects the circle $x^2+y^2-2x-2y-2=0$ orthogonally,passes through the point $(2,0)$,and touches the $X$-axis is:

If the coordinates of the point of contact of the circles $x^2+y^2-4x+8y+4=0$ and $x^2+y^2+2x=0$ are $(a, b)$,then $a+2b=$

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

The two circles $x^2 + y^2 - 2x - 3 = 0$ and $x^2 + y^2 - 4x - 6y - 8 = 0$ are such that:

$A$ circle $S$ cuts three circles $x^2+y^2-4x-2y+4=0$,$x^2+y^2-2x-4y+1=0$,and $x^2+y^2+4x+2y+1=0$ orthogonally. Then,the radius of $S$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module