Question

Find the radical centre of the three circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$

Answer 1

Let

$S_1:x^2+y^2-4x-6y+5=0$

$S_2:x^2+y^2-2x-4y-1=0$

$S_3:x^2+y^2-6x-2y=0$

Now, the radical axis of the circles will be

$S_1-S_2=0$

$(x^2+y^2-4x-6y+5)-(x^2+y^2-2x-4y-1)=0$

$2x+2y-6=0$

$\Rightarrow x+y=3.....\left(1\right)$

Similarly,

$S_2-S_3=0$

$(x^2+y^2-2x-4y-1)-(x^2+y^2-6x-2y)=0$

$2x-y=\frac{1}{2}.....\left(2\right)$

$S_3-S_1=0$

$(x^2+y^2-6x-2y)-(x^2+y^2-4x-6y+5)=0$

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

Adding equation $\left(1\right)\&\left(2\right)$, we have

$(x+y)+(2x-y)=3+\frac{1}{2}$

$x=\frac{7}{6}$

Substituting the value of $x$ in equation $\left(1\right)$, we have

$\frac{7}{6}+y=3$

$\Rightarrow y=3-\frac{7}{6}=\frac{11}{6}$

Since the value of $x$ and $y$ also satisfies equation $\left(3\right)$. Thus the radical centre of three circles is $(\frac{7}{6},\frac{11}{6})$.