Let the circle C_1: x^2+y^2-2(x+y)+1=0 and C_ | vedclass

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

Question

(A) The equation of circle $C_1$ is $x^2+y^2-2x-2y+1=0$. The centre of $C_1$ is $(1,1)$.The equation of circle $C_2$ with centre $(-1,0)$ and radius $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The equation of circle $C_1$ is $x^2+y^2-2x-2y+1=0$. The centre of $C_1$ is $(1,1)$.The equation of circle $C_2$ with centre $(-1,0)$ and radius $r=2$ is $(x+1)^2+(y-0)^2=2^2$,which simplifies to $x^2+2x+1+y^2=4$,or $x^2+y^2+2x-3=0$.The equation of the common chord is given by $S_1-S_2=0$:$(x^2+y^2-2x-2y+1) - (x^2+y^2+2x-3) = 0$$-4x-2y+4=0$$2x+y=2$.The line intersects the $y$-axis where $x=0$. Substituting $x=0$ into $2x+y=2$ gives $y=2$. Thus,point $P$ is $(0,2)$.The distance of $P(0,2)$ from the centre of $C_1(1,1)$ is $d = \sqrt{(1-0)^2+(1-2)^2} = \sqrt{1^2+(-1)^2} = \sqrt{2}$.The square of the distance is $d^2 = 2$.

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:

Similar Questions

The radius of the circle,having centre at $(2, 1)$ and one of its chords as a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,is

The length of the common chord of the circles $(x - a)^2 + y^2 = a^2$ and $x^2 + (y - b)^2 = b^2$ is

The equation of the pair of tangents drawn from the origin to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is:

The length of the common chord of the two circles $2x^{2} + 2y^{2} + 7x - 5y + 2 = 0$ and $x^{2} + y^{2} - 4x + 8y - 18 = 0$ is:

What is the angle subtended by the common chord of the circles $x^2 + y^2 - 4x - 4y = 0$ and $x^2 + y^2 = 16$ at the origin?

Vedclass Test Series

Exam Paper Generator

Online Exam Module