The common chord of the circles x^{2}+y^{2}-4 | vedclass

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

Question

(D) The given equations of the circles are $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$.Simplifying the second equation, we get $x^{2}+y^{2}=16$.The e

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The given equations of the circles are $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$.Simplifying the second equation, we get $x^{2}+y^{2}=16$.The equation of the common chord is obtained by subtracting the two equations: $(x^{2}+y^{2}-4x-4y) - (x^{2}+y^{2}-16) = 0$.This simplifies to $-4x-4y+16=0$, or $x+y=4$.Let the origin be $O(0,0)$. The circle $x^{2}+y^{2}-4x-4y=0$ passes through the origin. The common chord $x+y=4$ intersects the circle at points $A$ and $B$.Since the circle $x^{2}+y^{2}-4x-4y=0$ can be written as $(x-2)^{2}+(y-2)^{2}=8$, its center is $C(2,2)$ and radius is $r = \sqrt{8} = 2\sqrt{2}$.The distance from the center $C(2,2)$ to the line $x+y-4=0$ is $d = \frac{|2+2-4|}{\sqrt{1^{2}+1^{2}}} = 0$.Since the distance is $0$, the common chord is a diameter of the first circle.Any diameter subtends a right angle at any point on the circumference. Since the origin $(0,0)$ lies on the circle, the angle subtended by the diameter at the origin is $\frac{\pi}{2}$.

The common chord of the circles $x^{2}+y^{2}-4x-4y=0$ and $2x^{2}+2y^{2}=32$ subtends at the origin an angle equal to

Similar Questions

The lines represented by $5x^2-xy-5x+y=0$ are normals to a circle $S=0$. If this circle touches the circle $S^{\prime} \equiv x^2+y^2-2x+2y-7=0$ externally,then the equation of the chord of contact of the centre of $S^{\prime}=0$ with respect to $S=0$ is

Let the circle $C_1: x^2+y^2-2(x+y)+1=0$ and $C_2$ be a circle having centre at $(-1,0)$ and radius $2$. If the line of the common chord of $C_1$ and $C_2$ intersects the $y$-axis at the point $P$,then the square of the distance of $P$ from the centre of $C_1$ is:

If the circles $x^2+y^2-2x-2y-7=0$ and $x^2+y^2+4x+2y+k=0$ cut orthogonally,then the length of their common chord is units.

Tangents are drawn from the point $(4, 3)$ to the circle $x^2 + y^2 = 9$. The area of the triangle formed by them and the line joining their points of contact is

The distance between the chords of contact of the tangents to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ from the origin and the point $(g, f)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module