The angle at which the circles (x - 1)^2 + y^ | vedclass

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

Question

(B) The centers of the circles are $C_1 = (1, 0)$ and $C_2 = (0, 2)$,and their radii are $r_1 = \sqrt{10}$ and $r_2 = \sqrt{5}$.The distance between t

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) The centers of the circles are $C_1 = (1, 0)$ and $C_2 = (0, 2)$,and their radii are $r_1 = \sqrt{10}$ and $r_2 = \sqrt{5}$.The distance between the centers $d$ is given by $d = \sqrt{(1 - 0)^2 + (0 - 2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.The angle of intersection $	heta$ between two circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2 r_1 r_2}$.Substituting the values,we get $\cos 	heta = \frac{10 + 5 - 5}{2 \cdot \sqrt{10} \cdot \sqrt{5}} = \frac{10}{2 \cdot \sqrt{50}} = \frac{10}{2 \cdot 5\sqrt{2}} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = \frac{\pi}{4}$.

The angle at which the circles $(x - 1)^2 + y^2 = 10$ and $x^2 + (y - 2)^2 = 5$ intersect is

Similar Questions

Suppose $O(0,0)$ is the origin and the line $L = x + y - \lambda = 0$ meets the curve $x^2 + y^2 - 2x - 4y + 2 = 0$ at $A$ and $B$. If $\angle AOB = 90^{\circ}$,then the distance between such lines $L = 0$ is

$A$ common tangent to the circle $x^2+y^2=9$ and parabola $y^2=8x$ is

For a circle of diameter $R$,touching the circle $x^2 + y^2 - 4y = 0$ and passing through the point $(4, 5)$,which of the following is correct?

The equation of a common tangent to the circle $x^2+y^2=4$ and the ellipse $2x^2+25y^2=50$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module