Q.Show that the circles $x^2 + y^2 - 6x - 2y + 1 = 0$, $x^2 + y^2 + 2x - 8y + 13 = 0$ touch each other. Find the point of contact and the equation of common tangent at their point of contact.

Compare the distance between centres to the sum of the radii to confirm external contact, then find the point of contact by dividing the segment joining centres in the ratio $r_1:r_2$; the common tangent at that point is the radical axis $S_1-S_2=0$.

Circle 1: $x^2+y^2-6x-2y+1=0$, centre $C_1=(3,1)$, radius $r_1=\sqrt{3^2+1^2-1}=\sqrt9=3$.

Circle 2: $x^2+y^2+2x-8y+13=0$, centre $C_2=(-1,4)$, radius $r_2=\sqrt{(-1)^2+4^2-13}=\sqrt4=2$.

Distance between centres:

$C_1C_2=\sqrt{(3-(-1))^2+(1-4)^2}=\sqrt{16+9}=\sqrt{25}=5.$

Since $C_1C_2=5=r_1+r_2=3+2$, the circles touch each other externally.

Point of contact: it lies on segment $C_1C_2$ at distance $r_1$ from $C_1$, i.e. it divides $C_1C_2$ in the ratio $r_1:r_2=3:2$ from $C_1$:

$P=C_1+\frac{r_1}{C_1C_2}(C_2-C_1)=(3,1)+\frac35(-4,3)=\left(3-\frac{12}{5},\,1+\frac95\right)=\left(\frac35,\frac{14}{5}\right).$

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