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 equation of a circle passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.$(x^2+y^2+6x+4y-1

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of a circle passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.$(x^2+y^2+6x+4y-12) + \lambda(x^2+y^2-4x-6y-12) = 0$$(1+\lambda)x^2 + (1+\lambda)y^2 + (6-4\lambda)x + (4-6\lambda)y - 12(1+\lambda) = 0$Dividing by $(1+\lambda)$,we get $x^2 + y^2 + 2\left(\frac{3-2\lambda}{1+\lambda}\right)x + 2\left(\frac{2-3\lambda}{1+\lambda}\right)y - 12 = 0$.The radius $r$ of the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{g^2+f^2-c}$.Given $r = \sqrt{13}$,so $g^2+f^2-c = 13$.$\left(\frac{3-2\lambda}{1+\lambda}\right)^2 + \left(\frac{2-3\lambda}{1+\lambda}\right)^2 - (-12) = 13$$\frac{9+4\lambda^2-12\lambda + 4+9\lambda^2-12\lambda}{(1+\lambda)^2} = 1$$13\lambda^2 - 24\lambda + 13 = 1 + 2\lambda + \lambda^2$$12\lambda^2 - 26\lambda + 12 = 0 \Rightarrow 6\lambda^2 - 13\lambda + 6 = 0$.Solving for $\lambda$: $(2\lambda-3)(3\lambda-2) = 0$,so $\lambda = \frac{3}{2}$ or $\lambda = \frac{2}{3}$.For $\lambda = \frac{2}{3}$,the equation is $x^2+y^2+2x-12=0$.For $\lambda = \frac{3}{2}$,the equation is $x^2+y^2-2y-12=0$.

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+6x+4y-12=0$ and $x^2+y^2-4x-6y-12=0$ and having radius $\sqrt{13}$ is

Similar Questions

The equation of the circle passing through the point of intersection of the circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and the point $(3, -3)$ is:

The radical axis of the circles $3x^2 + 3y^2 - 7x + 8y + 11 = 0$ and $x^2 + y^2 - 3x - 4y + 5 = 0$ is

If $(\alpha, \beta)$ is the external centre of similitude of the circles $x^2+y^2=3$ and $x^2+y^2-2x+4y+4=0$,then $\frac{\beta}{\alpha}=$

From any point on the circle $x^2 + y^2 = a^2$,tangents are drawn to the circle $x^2 + y^2 = a^2 \sin^2 \alpha$. The angle between them is:

If $2 x^{2}+2 y^{2}+4 x+5 y+1=0$ and $3 x^{2}+3 y^{2}+6 x-7 y+3 k=0$ are orthogonal,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module