The angle between the circles x^2+y^2+4x-14y+ | vedclass

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

Question

(B) For the circle $x^2+y^2+2g_1x+2f_1y+c_1=0$,the center is $C_1(-g_1, -f_1)$ and radius $r_1 = \sqrt{g_1^2+f_1^2-c_1}$.For the first circle $x^2+y^2

Vedclass · Accepted Answer

$\cos^{-1} \left(\frac{3}{35}\right)$

Vedclass · Answer

(B) For the circle $x^2+y^2+2g_1x+2f_1y+c_1=0$,the center is $C_1(-g_1, -f_1)$ and radius $r_1 = \sqrt{g_1^2+f_1^2-c_1}$.For the first circle $x^2+y^2+4x-14y+28=0$,$g_1=2, f_1=-7, c_1=28$. Center $C_1(-2, 7)$,$r_1 = \sqrt{4+49-28} = \sqrt{25} = 5$.For the second circle $x^2+y^2-12x-6y-4=0$,$g_2=-6, f_2=-3, c_2=-4$. Center $C_2(6, 3)$,$r_2 = \sqrt{36+9-(-4)} = \sqrt{49} = 7$.The distance between centers $d = \sqrt{(6 - (-2))^2 + (3 - 7)^2} = \sqrt{8^2 + (-4)^2} = \sqrt{64+16} = \sqrt{80} = 4\sqrt{5}$.The angle $	heta$ between the circles is given by $\cos 	heta = \left| \frac{r_1^2+r_2^2-d^2}{2r_1r_2} ight|$.Substituting the values: $\cos 	heta = \left| \frac{25+49-80}{2 	imes 5 	imes 7} ight| = \left| \frac{74-80}{70} ight| = \left| \frac{-6}{70} ight| = \frac{3}{35}$.Therefore,$	heta = \cos^{-1} \left(\frac{3}{35}ight)$.

The angle between the circles $x^2+y^2+4x-14y+28=0$ and $x^2+y^2-12x-6y-4=0$ is

Similar Questions

If one end of the diameter of $x^2+y^2-2x-6y-15=0$ is $(4,1)$,then the coordinates of the other end are:

In the circle given below,let $OA = 1$ unit,$OB = 13$ units and $PQ \perp OB$. Then,the area of the triangle $PQB$ (in square units) is

Two circles $x^2 + y^2 = ax$ and $x^2 + y^2 = c^2$ touch each other if:

The angle between two circles,each passing through the centre of the other,is

The number of common tangents to the circles $x^2 + y^2 - 2x - 1 = 0$ and $x^2 + y^2 - 2y - 7 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module