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

The radical centre is the common point of the three radical axes, each found as $S_i-S_j=0$ for a pair of circles.

Let

$S_1: x^2+y^2-4x-6y+5=0,\quad S_2: x^2+y^2-2x-4y-1=0,\quad S_3: x^2+y^2-6x-2y=0.$

Radical axis of $S_1,S_2$: $S_1-S_2=0$:

$(-4x-6y+5)-(-2x-4y-1)=0 \implies -2x-2y+6=0 \implies x+y-3=0. \quad (i)$

Radical axis of $S_1,S_3$: $S_1-S_3=0$:

$(-4x-6y+5)-(-6x-2y)=0 \implies 2x-4y+5=0. \quad (ii)$

Solve $(i)$ and $(ii)$: from $(i)$, $x=3-y$. Substitute into $(ii)$:

$2(3-y)-4y+5=0 \implies 6-2y-4y+5=0 \implies 11=6y \implies y=\frac{11}{6}.$

$x=3-\frac{11}{6}=\frac{7}{6}.$

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