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

The radical axis of the circles $S_1: x^2+y^2-4x+6y-10=0$ and $S_2: x^2+y^2+2x-6y+2=0$ cuts the circle $S_1$ in

The radical centre of the circles $x^2 + y^2 + 4x + 6y = 19$,$x^2 + y^2 = 9$,and $x^2 + y^2 - 2x - 2y = 5$ is:

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k$ is equal to

The equation of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $2{x^2} + 2{y^2} - 10x - 12y + 12 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module