If the radical centre of the three circles x^ | vedclass

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

Question

(D) Let the equations of the circles be $S_1: x^2+y^2-1=0$,$S_2: x^2+y^2-2x-3=0$,and $S_3: x^2+y^2-2y-3=0$. To find the radical centre,we find the int

Vedclass · Accepted Answer

$(x+1)^2+(y+1)^2=25$

Vedclass · Answer

(D) Let the equations of the circles be $S_1: x^2+y^2-1=0$,$S_2: x^2+y^2-2x-3=0$,and $S_3: x^2+y^2-2y-3=0$. To find the radical centre,we find the intersection of the radical axes. $S_1-S_2=0 \implies (x^2+y^2-1) - (x^2+y^2-2x-3) = 0 \implies 2x+2=0 \implies x=-1$. $S_1-S_3=0 \implies (x^2+y^2-1) - (x^2+y^2-2y-3) = 0 \implies 2y+2=0 \implies y=-1$. Thus,the radical centre $C(\alpha, \beta)$ is $(-1, -1)$. The radii of the circles are: $r_1 = \sqrt{0^2+0^2-(-1)} = 1$. $r_2 = \sqrt{1^2+0^2-(-3)} = \sqrt{4} = 2$. $r_3 = \sqrt{0^2+1^2-(-3)} = \sqrt{4} = 2$. The sum of the radii is $r = 1+2+2 = 5$. The equation of the circle with centre $(-1, -1)$ and radius $5$ is $(x-(-1))^2 + (y-(-1))^2 = 5^2$,which simplifies to $(x+1)^2+(y+1)^2=25$.

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is:

Similar Questions

If the circle $S=0$ intersects the three circles $S_1 \equiv x^2+y^2+4x-7=0$,$S_2 \equiv x^2+y^2+y=0$ and $S_3 \equiv x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$ orthogonally,then the radical axis of $S=0$ and $S_1=0$ is

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :

If the circle $S \equiv x^2+y^2+2gx+4y+1=0$ bisects the circumference of the circle $x^2+y^2-2x-3=0$,then the radius of circle $S=0$ is

$P$ is a point of intersection of the circles $S \equiv x^2+y^2-6x+2ky+1=0$ and $S' \equiv x^2+y^2+2kx-6y-7=0$. If the tangent at $P$ to $S=0$ passes through the centre of $S'=0$ and the tangent at $P$ to $S'=0$ passes through the centre of $S=0$,then the radius of $S'=0$ is

The length of the tangent drawn from any point on the circle $x^2+y^2+2gx+2fy+c_1=0$ to the circle $x^2+y^2+2gx+2fy+c_2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module