Question

The vertices of a triangle are $A(2,-1),B(3,1)$ and $C(1,-2).$ Three circles are drawn with $AB,BC,CA$ as diameters. If a circle $S=0$ intersects all the three circles orthogonally, then the radius of $S=0$ is equal to

• A. $\sqrt{125}$
• B. $\sqrt{120}$
• C. $\sqrt{84}$
• D. $\sqrt{45}$

Answer 1

The correct option is B $\sqrt{120}$

Given vertices : $A(2,-1),B(3,1)$ and $C(1,-2).$
Given $S=0$ intersects all the three circles having $AB,BC,CA$ as diameters.
So, centre of $S=0$ is the radical centre of these three circles which is the orthocentre of the $△ABC.$

Equation of altitude through $A:$
$y+1=-\frac{2}{3}(x-2)\Rightarrow 2x+3y-1=0\dots \left(1\right)$
Equation of altitude through $B:$
$y-1=-1(x-3)\Rightarrow x+y-4=0\dots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$ we get radical centre $\equiv (11,-7)$

Equation of circle having $AB$ as a diameter is
$(x-2)(x-3)+(y+1)(y-1)=0\Rightarrow x^2+y^2-5x+5=0$

Now, radius of $S=0$ is equal to length of tangent from $(11,-7)$ to $x^2+y^2-5x+5=0$
$\therefore$ radius of $S=0$ is $\sqrt{121+49-55+5}=\sqrt{120}$