Number of circles intersecting x^2+y^2=4,x^2+ | vedclass

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

Question

(A) Let the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$. For a circle to be orthogonal to $x^2+y^2+2g_i x+2f_i y+c_i=0$,the condition is $2gg_i+

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$. For a circle to be orthogonal to $x^2+y^2+2g_i x+2f_i y+c_i=0$,the condition is $2gg_i+2ff_i=c+c_i$. Given circles: $S_1: x^2+y^2-4=0 \Rightarrow g_1=0, f_1=0, c_1=-4$ $S_2: x^2+y^2-2x-3=0 \Rightarrow g_2=-1, f_2=0, c_2=-3$ $S_3: x^2+y^2-2y-3=0 \Rightarrow g_3=0, f_3=-1, c_3=-3$ Applying the condition: $1$) For $S_1$: $2g(0)+2f(0)=c-4 \Rightarrow c=4$. $2$) For $S_2$: $2g(-1)+2f(0)=c-3$ $\Rightarrow -2g=4-3=1$ $\Rightarrow g=-1/2$. $3$) For $S_3$: $2g(0)+2f(-1)=c-3$ $\Rightarrow -2f=4-3=1$ $\Rightarrow f=-1/2$. Now,check the radius $r$ of the circle $S$: $r^2 = g^2+f^2-c = (-1/2)^2+(-1/2)^2-4 = 1/4+1/4-4 = 1/2-4 = -7/2$. Since $r^2 Thus,the number of such circles is $0$.

Number of circles intersecting $x^2+y^2=4$,$x^2+y^2-2x-3=0$ and $x^2+y^2-2y-3=0$ orthogonally is

Similar Questions

If the circles $x^2+y^2-6x-8y-12=0$ and $x^2+y^2-4x+6y+k=0$ are orthogonal to each other,then the value of $k$ is:

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

If the origin lies on a diameter of the circle $x^2+y^2-4x-2y-4=0$,then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is

Let the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ cut the circles $x^2+y^2-2x+2y-2=0$ and $x^2+y^2+4x-6y+9=0$ orthogonally. If the centre of the circle $S=0$ lies on the line $2x+3y-2=0$,then $2g+f=$

The equation of the line perpendicular to the radical axis of two circles $x^2+y^2-5x+6y+12=0$ and $x^2+y^2+6x-4y-14=0$,and passing through $(1,1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module