Let theta be the angle between the circles S | vedclass

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

Question

(A) For circle $S: x^2+y^2+2x-2y+c=0$,the center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - c} = \sqrt{2-c}$.For circle $S': x^2+y^2-6x-8y

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) For circle $S: x^2+y^2+2x-2y+c=0$,the center $C_1 = (-1, 1)$ and radius $r_1 = \sqrt{(-1)^2 + 1^2 - c} = \sqrt{2-c}$.For circle $S': x^2+y^2-6x-8y+9=0$,the center $C_2 = (3, 4)$ and radius $r_2 = \sqrt{3^2 + 4^2 - 9} = \sqrt{16} = 4$.The distance between centers $d = \sqrt{(3 - (-1))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = 5$.The angle $	heta$ between two circles is given by $\cos 	heta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$.Substituting the values: $\frac{5}{16} = \frac{5^2 - (2-c) - 4^2}{2 \cdot \sqrt{2-c} \cdot 4} = \frac{25 - 2 + c - 16}{8\sqrt{2-c}} = \frac{7+c}{8\sqrt{2-c}}$.Simplifying: $\frac{5}{16} = \frac{7+c}{8\sqrt{2-c}} \implies \frac{5}{2} = \frac{7+c}{\sqrt{2-c}}$.Squaring both sides: $\frac{25}{4} = \frac{(7+c)^2}{2-c} \implies 25(2-c) = 4(49 + 14c + c^2)$.$50 - 25c = 196 + 56c + 4c^2 \implies 4c^2 + 81c + 146 = 0$.Solving for $c$: $c = \frac{-81 \pm \sqrt{81^2 - 4(4)(146)}}{8} = \frac{-81 \pm \sqrt{6561 - 2336}}{8} = \frac{-81 \pm \sqrt{4225}}{8} = \frac{-81 \pm 65}{8}$.$c_1 = \frac{-16}{8} = -2$ and $c_2 = \frac{-146}{8} = -18.25$.Since $c$ is an integer,$c = -2$.The radius $r_1 = \sqrt{2 - (-2)} = \sqrt{4} = 2$.

Let $\theta$ be the angle between the circles $S \equiv x^2+y^2+2x-2y+c=0$ and $S' \equiv x^2+y^2-6x-8y+9=0$. If $c$ is an integer and $\cos \theta = \frac{5}{16}$,then the radius of the circle $S=0$ is

Similar Questions

The image of the point $(3, 4)$ with respect to the radical axis of the circles $x^2 + y^2 + 8x + 2y + 10 = 0$ and $x^2 + y^2 + 7x + 3y + 10 = 0$ is

If the equation of the incircle of an equilateral triangle is $x^2 + y^2 + 4x - 6y + 4 = 0$,then the equation of its circumcircle is:

If the radical axis of the circles $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ forms a triangle of area $A$ with the coordinate axes,then the value of $\frac{1}{A}$ is

$A$ circle $C_1$ of radius $2$ touches both $x$-axis and $y$-axis. Another circle $C_2$ whose radius is greater than $2$ touches circle $C_1$ and both the axes. Then the radius of circle $C_2$ is

The equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2+y^2-6x-4y-12=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module