Q.Find the transverse common tangents of the circles $x^2+y^2-4x-10y+28=0$ and $x^2+y^2+4x-6y+4=0$.

Find each circle's centre/radius; since the distance between centres exceeds the sum of the radii, transverse common tangents exist and pass through the internal centre of similitude, which divides the join of centres in the ratio $r_1:r_2$.

Circle 1: $x^2+y^2-4x-10y+28=0$: centre $C_1=(2,5)$, $r_1=\sqrt{4+25-28}=1$

Circle 2: $x^2+y^2+4x-6y+4=0$: centre $C_2=(-2,3)$, $r_2=\sqrt{4+9-4}=3$

$C_1C_2=\sqrt{(2-(-2))^2+(5-3)^2}=\sqrt{16+4}=2\sqrt5\approx4.47$

Since $C_1C_2>r_1+r_2\,(=4)$, the circles are separate and two transverse (internal) common tangents exist, meeting at the internal centre of similitude $I$, which divides $C_1C_2$ internally in the ratio $r_1:r_2=1:3$:

$I=\left(\dfrac{1(-2)+3(2)}{4},\dfrac{1(3)+3(5)}{4}\right)=\left(1,\dfrac92\right)$

A line through $I$: $y-\dfrac92=m(x-1)$, i.e. $mx-y+\left(\dfrac92-m\right)=0$.

Distance from $C_1=(2,5)$ equal to $r_1=1$:

$\dfrac{\left|2m-5+\tfrac92-m\right|}{\sqrt{m^2+1}}=1 \Rightarrow \left|m-\dfrac12\right|=\sqrt{m^2+1}$

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