Find the angle of intersection of the two cir | vedclass

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

Question

(D) For the first circle $x^2 + y^2 - 2x - 2y = 0$,the center $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}$.For the second circle $x^2

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) For the first circle $x^2 + y^2 - 2x - 2y = 0$,the center $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}$.For the second circle $x^2 + y^2 = 4$,the center $C_2 = (0, 0)$ and radius $r_2 = 2$.The distance between the centers $d = C_1C_2 = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{2}$.If $	heta$ is the angle of intersection,then $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$.Substituting the values,$\cos 	heta = \frac{2 + 4 - 2}{2 	imes \sqrt{2} 	imes 2} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = 45^\circ$.

Find the angle of intersection of the two circles $x^2 + y^2 - 2x - 2y = 0$ and $x^2 + y^2 = 4$ in degrees.

Similar Questions

Let two circles $C_1$ and $C_2$ of radii $r_1 = 2$ and $r_2 = 4$ be tangent to each other at point $P$ and tangent to a common straight line (not passing through $P$) at points $Q$ and $R$ respectively. Then the value of $PQ^2 + QR^2 + RP^2$ is:

Let the normals at all the points on a given curve pass through a fixed point $(a, b)$. If the curve passes through $(3, -3)$ and $(4, -2\sqrt{2})$,and given that $a - 2\sqrt{2}b = 3$,then $(a^{2} + b^{2} + ab)$ is equal to ..... .

If the circles $x^2+y^2=9$ and $x^2+y^2+2\alpha x+2y+1=0$ touch each other internally,then the value of $\alpha^3$ is

The circle touching the coordinate axes with its centre lying on $x-2y-3=0$ is

The ratio of the largest and shortest distances from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module