The angle between circles x^2+y^2+2x+4y+1=0 a | vedclass

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

Question

(A) The given circles are $C_1: x^2+y^2+2x+4y+1=0$ and $C_2: x^2+y^2-2x+6y-3=0$.Comparing with the general form $x^2+y^2+2gx+2fy+c=0$:For $C_1$: $g_1=

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$

Vedclass · Answer

(A) The given circles are $C_1: x^2+y^2+2x+4y+1=0$ and $C_2: x^2+y^2-2x+6y-3=0$.Comparing with the general form $x^2+y^2+2gx+2fy+c=0$:For $C_1$: $g_1=1, f_1=2, c_1=1$. Centre $O_1=(-1, -2)$,radius $r_1=\sqrt{1^2+2^2-1}=2$.For $C_2$: $g_2=-1, f_2=3, c_2=-3$. Centre $O_2=(1, -3)$,radius $r_2=\sqrt{(-1)^2+3^2-(-3)}=\sqrt{1+9+3}=\sqrt{13}$.The distance $d$ between the centres $O_1(-1, -2)$ and $O_2(1, -3)$ is $d=\sqrt{(1-(-1))^2+(-3-(-2))^2}=\sqrt{2^2+(-1)^2}=\sqrt{5}$.The angle $	heta$ between the circles is given by $\cos 	heta = \left|\frac{d^2-r_1^2-r_2^2}{2r_1r_2}ight|$.Substituting the values: $\cos 	heta = \left|\frac{5-4-13}{2 	imes 2 	imes \sqrt{13}}ight| = \left|\frac{-12}{4\sqrt{13}}ight| = \frac{3}{\sqrt{13}}$.Therefore,$	heta = \cos^{-1}\left(\frac{3}{\sqrt{13}}ight)$.

The angle between circles $x^2+y^2+2x+4y+1=0$ and $x^2+y^2-2x+6y-3=0$ is

Similar Questions

The radical axis of the circles $S_1: x^2+y^2-4x+6y-10=0$ and $S_2: x^2+y^2+2x-6y+2=0$ cuts the circle $S_1$ in

The equation of the circle described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as diameter is:

If the circles $x^2+y^2+2kx-4y+1=0$ and $x^2+y^2-8x-12y+43=0$ touch each other,then $k=$

The equation of the circle having the chord $x \cos \alpha + y \sin \alpha = p$ of the circle $x^2 + y^2 = a^2$ as its diameter is:

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module