Q.Find the radical axis of the circles $x^2+y^2-2x-4y-6=0$ and $x^2+y^2+4x-2y+1=0$.

Subtracting the two circle equations gives the common chord $S_1-S_2=0$, i.e. $2x+1=0$.

For two circles $S_1=0$ and $S_2=0$ (each with unit coefficients of $x^2,y^2$), the common chord is $S_1-S_2=0$.

$S_1=x^2+y^2+2x+3y+1,\qquad S_2=x^2+y^2+4x+3y+2.$

$S_1-S_2=(2x-4x)+(3y-3y)+(1-2)=-2x-1.$

Setting this to zero:

$-2x-1=0\ \Rightarrow\ 2x+1=0.$

Normalising the second circle and subtracting gives the radical axis $x + 10y - 13 = 0$.

The radical axis is obtained by subtracting the two circle equations, each written with coefficient $1$ for $x^2$ and $y^2$.

$S_1: x^2 + y^2 - 3x - 4y + 5 = 0$.

$S_2: 3(x^2+y^2) - 7x + 8y - 11 = 0 \Rightarrow x^2 + y^2 - \tfrac73 x + \tfrac83 y - \tfrac{11}{3} = 0$.

Radical axis $S_1 - S_2 = 0$:

$\left(-3 + \tfrac73\right)x + \left(-4 - \tfrac83\right)y + \left(5 + \tfrac{11}{3}\right) = 0$

$-\tfrac23 x - \tfrac{20}{3} y + \tfrac{26}{3} = 0$.

Multiply by $-\tfrac32$:

$x + 10y - 13 = 0$.

When two circles touch, the common tangent at the point of contact is their radical axis, $S_1-S_2=0$.

Circle 1: $x^2+y^2+10x-2y+22=0 \Rightarrow g_1=5,f_1=-1,c_1=22$, $r_1=\sqrt{25+1-22}=2$, $C_1=(-5,1)$.

Circle 2: $x^2+y^2+2x-8y+8=0 \Rightarrow g_2=1,f_2=-4,c_2=8$, $r_2=\sqrt{1+16-8}=3$, $C_2=(-1,4)$.

Distance between centres: $d=\sqrt{(-5-(-1))^2+(1-4)^2}=\sqrt{16+9}=5=r_1+r_2$, so the circles touch externally, confirming a common tangent exists at the point of contact.

The radical axis (obtained by subtracting the two equations) is this common tangent:

$(10x-2y+22)-(2x-8y+8)=0$

$8x+6y+14=0$

$4x+3y+7=0$

For two intersecting circles $S_1=0$ and $S_2=0$, the common chord is the line $S_1-S_2=0$ (the $x^2,y^2$ terms cancel, leaving a linear equation).

$S_1 \equiv x^2+y^2-4x-4y+3=0$

$S_2 \equiv x^2+y^2-5x-6y+4=0$

$S_1-S_2$:

$(-4x-4y+3)-(-5x-6y+4)$

$= (-4x+5x)+(-4y+6y)+(3-4)$

$= x+2y-1$

So $S_1-S_2=0$ gives the common chord.

The radical axis of two circles $S_1=0$ and $S_2=0$ (each with unit coefficient of $x^2+y^2$) is $S_1-S_2=0$.

$S_1: x^2+y^2+4x+6y-7=0$

The second circle is $4(x^2+y^2)+8x+12y-9=0$; divide by 4 to make the coefficient of $x^2+y^2$ equal to 1:

$S_2: x^2+y^2+2x+3y-\frac{9}{4}=0$

Radical axis: $S_1-S_2=0$

$(4x+6y-7)-\left(2x+3y-\frac{9}{4}\right)=0$

$2x+3y-7+\frac{9}{4}=0$

$2x+3y-\frac{19}{4}=0$

Multiply by 4:

$8x+12y-19=0$