Question

Let $S$ be the circle which cuts the three circles $x^2+y^2-3x-6y+14=0,x^2+y^2-x-4y+8=0$ and $x^2+y^2+2x-6y+9=0$ orthogonally. Then

• A. From the point $(2,3)$, two tangents can be drawn to $S$
• B. The radius of $S$ is $2$
• C. The absolute difference between the coordinates of the center of $S$ is $1$
• D. The center of $S$ is $(1,2)$

Answer 1

Answer: (B), (C), (D)

The correct options are
B The radius of $S$ is $2$
C The absolute difference between the coordinates of the center of $S$ is $1$
D The center of $S$ is $(1,2)$

The circle having center at the radical center of the three given circles and radius equal to the length of the tangent from it to any one of the three circles cuts all the three circles orthogonally. The given circles are

$x^2+y^2-3x-6y+14=0,....\left(1\right)x^2+y^2-x-4y+8=0,....\left(2\right)x^2+y^2+2x-6y+9=0....\left(3\right)$

The radical axes of $\left(1\right),\left(2\right)\text{and}\left(3\right)$ are, respectively.

$x+y-3=0.....\left(4\right)3x-2y+1=0.....\left(5\right)$

Solving $\left(4\right)\text{and}\left(5\right)$, we get $x=1,y=2$
Thus, the coordinates of the radical center are $(1,2)$. The length of the tangent from $(1,2)$ to $\left(1\right)$ is
$r=\sqrt{1+4-3-12+14}=2$

Hence, the equation of the required circle is
$(x-1{)}^2+(y-2{)}^2=2^2\Rightarrow x^2+y^2-2x-4y+1=0$