Q.Find the common tangent of the circles $x^2 + y^2 + 10x - 2y + 22 = 0$ and $x^2 + y^2 + 2x - 8y + 8 = 0$ at their point of contact.

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 $S_1-S_2=0$.

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

…

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 $S_1 - S_2 = 0$:

…

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

…

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).

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

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

$S_1-S_2$:

$(-4x-4y+3)-(-5x-6y+4)$ …

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: $S_1-S_2=0$

…