If the acute angle between the circles S equi | vedclass

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

Question

(A) The circles are $S: x^2+y^2+2kx+4y-3=0$ and $S': x^2+y^2-4x+2ky+9=0$. The centres are $C_1 = (-k, -2)$ and $C_2 = (2, -k)$. The radii are $r_1 = \

Vedclass · Accepted Answer

$x-5y+6=0$

Vedclass · Answer

(A) The circles are $S: x^2+y^2+2kx+4y-3=0$ and $S': x^2+y^2-4x+2ky+9=0$. The centres are $C_1 = (-k, -2)$ and $C_2 = (2, -k)$. The radii are $r_1 = \sqrt{k^2+4+3} = \sqrt{k^2+7}$ and $r_2 = \sqrt{4+k^2-9} = \sqrt{k^2-5}$. The distance between centres is $d^2 = (2+k)^2 + (-k+2)^2 = 4+4k+k^2 + k^2-4k+4 = 2k^2+8$. The angle $	heta$ between circles satisfies $\cos 	heta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$. Given $\cos 	heta = \frac{3}{8}$,so $\frac{2k^2+8-(k^2+7)-(k^2-5)}{2\sqrt{k^2+7}\sqrt{k^2-5}} = \frac{3}{8}$. $\frac{6}{2\sqrt{(k^2+7)(k^2-5)}} = \frac{3}{8} \implies \frac{3}{\sqrt{k^4+2k^2-35}} = \frac{3}{8}$. $\sqrt{k^4+2k^2-35} = 8 \implies k^4+2k^2-35 = 64 \implies k^4+2k^2-99 = 0$. $(k^2+11)(k^2-9) = 0$. Since $k^2 > 0$,$k^2=9$,so $k=3$ or $k=-3$. The centre of $S'=0$ is $(2, -k)$. For it to be in the first quadrant,$-k > 0$,so $k=-3$. The radical axis is $S-S'=0$: $(2k+4)x + (4-2k)y - 12 = 0$. Substituting $k=-3$: $(2(-3)+4)x + (4-2(-3))y - 12 = 0 \implies -2x + 10y - 12 = 0 \implies x-5y+6=0$.

If the acute angle between the circles $S \equiv x^2+y^2+2kx+4y-3=0$ and $S' \equiv x^2+y^2-4x+2ky+9=0$ is $\cos^{-1}(\frac{3}{8})$ and the centre of $S'=0$ lies in the first quadrant,then the radical axis of $S=0$ and $S'=0$ is

Similar Questions

The point/points of intersection of the common tangents of the two circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2y-15=0$ is/are

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 curves $x^{2}-6x+y^{2}+8=0$ and $x^{2}-8y+y^{2}+16-k=0$ $(k>0)$ touch each other at a point,then the largest value of $k$ is

If the circles $x^2+y^2+kx+4y+2=0$ and $2(x^2+y^2)-4x-3y+k=0$ cut orthogonally,then $k=$

If the centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$,then $2\alpha+\beta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module