Question

Consider the circles $S_1:x^2+y^2+2x+4y+1=0,S_2:x^2+y^2-4x+3=0\&S_3:x^2+y^2+6y+5=0$ which of this following statements are correct ?

• A. Radical centre of $S_1,S_2$ and $S_3$ lies in $1^{st}$ quadrant
• B. Radical centre of $S_1,S_2$ and $S_3$ lies in $4^{th}$ quadrant
• C. Radius of the circle intersecting $S_1,S_2$ and $S_3$ orthogonally is 1.
• D. Circle orthogonal to $S_1,S_2$ and $S_3$ has its x and y intercept equal to zero.

Answer 1

Answer: (B), (C)

The correct options are
B Radical centre of $S_1,S_2$ and $S_3$ lies in $4^{th}$ quadrant
C Radius of the circle intersecting $S_1,S_2$ and $S_3$ orthogonally is 1.

Radical axes of two circles $S_1$ and $S_2$ is given by $S_1-S_2=0.$

So, for given circles radical axes are,

$6x+4y-2=0...\left(1\right)$

$2y-2x+4=0...\left(2\right)$

$4x+6y+2=\mathrm{0...}\left(3\right)$

Let's find the radical center by finding intersection of the lines,

(1) - 3(2)

$\to 10y=-10\to y=-1$

Now,
$6x+4y-2=0$

$6x-4-2=0$

$\to x=1$

Hence, the radical center is (1.-1).

So, the option b is correct answer.

We know that radical center is center of the circle orthogona to three circles.

And also we know that length of tangent from center of orthogonal circle to any of parent circle is radius of orthogonal circle.

Now length of tangent from pont (h,k) to circle $x^2+y^2+2gx+2fy+c=0$ is given by

$\sqrt{h^2+k^2+2gh+2fk+c=0}$

So, length of tangent from center of orthogonal circle i.e., (1,-1) to circle $x^2+y^2+2x+4y+1=0$ is,

$=\sqrt{1^2+(-1{)}^2+2(1)+4(-1)+1}$

$=1$

$=$ Radius of orthogonal circle

So, C is also the correct answer.

Now, we know that the general equation of circle with center (h,k) and radius r is,

$\to x^2+y^2+2gx+2fy+c=0$

So, for orthogonal circle, equation is,

$\to (x-1{)}^2+(y+1{)}^2=1$

So, the intersection points of circle with x-axis are $(2,0),(0,0)\to x-intercept=2$

And, the intersection points of circle with y-axis are $(0,0),(0,-2)\to y-intercept=2$

Hence, (B) and (C) are the correct answer.