Question

Consider three circles whose equations are $x^2+y^2+3x+2y+1=0$, $x^2+y^2-x+6y+5=0$ and $x^2+y^2+5x-8y+15=0$, then

• A. radical centre of the circles is $(3,2)$
• B. equation of the circle which is orthogonal to given circles is$x^2+y^2-6x-4y-14=0$
• C. equation of the circle which is orthogonal to given circles is$x^2+y^2-6x+4y-13=0$
• D. radical centre of the circles is$(3,-2)$

Answer 1

Answer: (A), (B)

The correct options are
A radical centre of the circles is $(3,2)$
B equation of the circle which is orthogonal to given circles is$x^2+y^2-6x-4y-14=0$

Given circles are
$S_1:x^2+y^2+3x+2y+1=0$
$S_2:x^2+y^2-x+6y+5=0$
$S_3:x^2+y^2+5x-8y+15=0$

Equations of two radical axes are
$S_1-S_2=0\Rightarrow 4x-4y-4=0\Rightarrow x-y-1=0$

And
$S_2-S_3=0\Rightarrow -6x+14y-10=0\Rightarrow 3x-7y+5=0$

Solving them, we get the radical centre as $(3,2)$
Also, if $r$ is the length of the tangent drawn from the radical centre $(3,2)$ to any one of the given circles, say $S_1$, we have
$r=\sqrt{S_1}=\sqrt{3^2+2^2+3.3+2.2+1}\Rightarrow r=\sqrt{27}$

Hence, $(3,2)$ is the centre and $\sqrt{27}$ is the radius of the circle intersecting them orthogonally.
$\therefore$ Its equation is
$(x-3{)}^2+(y-2{)}^2=27\Rightarrow x^2+y^2-6x-4y-14=0$