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

If $A\left(\frac{\pi}{3}\right)$ and $B\left(\frac{\pi}{6}\right)$ are points on a circle represented in parametric form with center $(0,0)$ and radius $12$,then the length of the chord $AB$ is:

If the four distinct points $(4,6), (-1,5), (0,0)$ and $(k, 3k)$ lie on a circle of radius $r$,then $10k + r^2$ is equal to

$A$ circle is drawn in a sector of a larger circle of radius $r$,as shown in the figure. The smaller circle is tangent to the two bounding radii and the arc of the sector. The radius of the small circle is

Suppose that the earth is a sphere of radius $6400 \, km$. The height from the earth's surface from where exactly a fourth of the earth's surface is visible,is $...... \, km$.

Let $A(1, 4)$ and $B(1, -5)$ be two points. Let $P$ be a point on the circle $(x-1)^{2} + (y-1)^{2} = 1$ such that $(PA)^{2} + (PB)^{2}$ has a maximum value. Then the points $P, A,$ and $B$ lie on:

Vedclass Test Series

Exam Paper Generator

Online Exam Module