The point (3, -4) lies on both the circles x^ | vedclass

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

Question

(D) Given circles are $S_1: x^2 + y^2 - 2x + 8y + 13 = 0$ and $S_2: x^2 + y^2 - 4x + 6y + 11 = 0$. For $S_1$,the center $C_1 = (1, -4)$ and radius $r_

Vedclass · Accepted Answer

$135^{\circ}$

Vedclass · Answer

(D) Given circles are $S_1: x^2 + y^2 - 2x + 8y + 13 = 0$ and $S_2: x^2 + y^2 - 4x + 6y + 11 = 0$. For $S_1$,the center $C_1 = (1, -4)$ and radius $r_1 = \sqrt{1^2 + (-4)^2 - 13} = \sqrt{1 + 16 - 13} = 2$. For $S_2$,the center $C_2 = (2, -3)$ and radius $r_2 = \sqrt{2^2 + (-3)^2 - 11} = \sqrt{4 + 9 - 11} = \sqrt{2}$. The distance between the centers $d = C_1C_2 = \sqrt{(2 - 1)^2 + (-3 - (-4))^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$. The angle $	heta$ between the circles is given by $\cos 	heta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$. Substituting the values: $\cos 	heta = \frac{(\sqrt{2})^2 - 2^2 - (\sqrt{2})^2}{2 	imes 2 	imes \sqrt{2}} = \frac{2 - 4 - 2}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$. Thus,$	heta = 135^{\circ}$.

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

Similar Questions

If the lengths of the tangents drawn from a point $P$ to the three circles $x^2+y^2-4=0$,$x^2+y^2-2x+3y=0$,and $x^2+y^2+7y-18=0$ are equal,then the coordinates of $P$ are

The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve

If the equation of the circle which cuts each of the circles $x^2+y^2=4$,$x^2+y^2-6x-8y+10=0$,and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2gx+2fy+c=0$,then $g+f+c=$

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module