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 the two circles' centres/radii, locate the internal centre of similitude (divides the centre-line in ratio $r_1:r_2$ internally), then find tangents from that point.

Circle 1: $x^2+y^2-4x-10y+28=0$, centre $C_1=(2,5)$, $r_1=\sqrt{4+25-28}=\sqrt1=1$

Circle 2: $x^2+y^2+4x-6y+4=0$, centre $C_2=(-2,3)$, $r_2=\sqrt{4+9-4}=\sqrt9=3$

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

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

$P=\left(\dfrac{1(-2)+3(2)}{1+3},\ \dfrac{1(3)+3(5)}{1+3}\right)=\left(\dfrac{4}{4},\dfrac{18}{4}\right)=\left(1,\dfrac92\right)$

Now find the lines through $P(1,\frac92)$ tangent to circle 1 (centre $(2,5)$, radius 1). Let the line be $m(x-1)-(y-\frac92)=0$, i.e. $mx-y+\left(\frac92-m\right)=0$.

Distance from $(2,5)$ equals 1:

$\dfrac{\left|2m-5+\frac92-m\right|}{\sqrt{m^2+1}}=1 \Rightarrow \left|m-\dfrac12\right|=\sqrt{m^2+1}$

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