Question

Find all the common tangents to the circles
$x^2+y^2-2x-6y+9=0$
and $x^2+y^2+6x-2y+1=0$.

Answer 1

The centres and radii of the circles are
$C_1(1,3)$ and $r_1=\sqrt{1+9-9}=1$.
$C_2(-3,1)$ and $r_2=\sqrt{9+1-1}=3$.
$C_1{C}_2=\sqrt{20},r_1+r_2=4=\sqrt{16}$
$\therefore C_1{C}_2>r_1+r_2$. Hence the circles are non-intersecting. Thus there will be four common tangents.
Transverse common tangents are tangents drawn from the point $P$ which divides $C_1{C}_2$ internally in the ratio of radii $1:3$.
Co-ordinates of $P$ are
$(\frac{1(-3)+3.1}{1+3},\frac{1.1+3.3}{1+3})$ i.e. $(0,\frac{5}{2})$.
Direct common tangents are tangents drawn from the point $Q$ which divides $C_1{C}_2$ externally in the ratio $1:3$.
Co-ordinates of $Q$ are tangents through the point $P(0,5/2)$.
Any line through $(0,5/2)$ is
$y-5/2=mx.....\left(1\right)$
or $mx-y+5/2=0$.
Apply the usual condition of tangency to any of the circle
$\therefore \frac{m.1-3+5/2}{\sqrt{(m^2+1)}}=1$
or ${(m-\frac{1}{2})}^2=m^2+1$
or $-m-3/4=0$ or $0m^2-m-3/4=0$.
Hence $m=-3/4$ and $\infty$ as coeff. of $m^2$ is zero.
Therefore from $\left(1\right)$,
$\frac{y-5/2}{x}=m=\infty$ and $-3/4$.
$\therefore x=0$ is a tangent and $y-5/2=-3x/4$
or $3x+4y-10=0$ is another tangent.
Direct tangents are tangents drawn from the point $Q(3,4)$.
Now proceeding as for transverse tangents their equations are
$y=4,4x-3y=0$.