Question

Find all the common tangents to the circle
\[
  {x^2} + {y^2} - 2x - 6y + 9 = 0 \\
  {x^2} + {y^2} + 6x - 2y + 1 = 0 \\
  \]

Answer 1

Hint: First of all calculate the condition among the centres of the circle and their radius , as per the given equation then calculate their direct common tangent and transverse common tangent.

Complete step-by-step answer:

Given the equation of circles as
\[
  {x^2} + {y^2} - 2x - 6y + 9 = 0 \\
  {x^2} + {y^2} + 6x - 2y + 1 = 0 \\
  \]
First we convert them into standard form of circle which is \[{(x - a)^2} + {(y - b)^2} = {r^2}\], so we get,
\[
  {(x - 1)^2} + {(y - 3)^2} = {1^2} \\
  {(x - ( - 3))^2} + {(y - 1)^2} = {3^2} \\
  \]
The centres and radii of the circles are
\[
  {c_1} = (1,3),{r_1} = 1 \\
  {c_2} = ( - 3,1),{r_2} = 3 \\
  \]
The distance among centres and sum of radii of the circle are given as
\[
  {c_1}{c_2} = \sqrt {{{( - 4)}^2} + {{(2)}^2}} = \sqrt {20} \\
  {r_1} + {r_2} = 1 + 3 = 4 \\
  \]
As \[{c_1}{c_2} > {r_1} + {r_2}\]so there are four possible tangents.
Transverse common tangents are tangents drawn from the point P which divides \[{c_1}{c_2}\] ​ internally in the ratio of radii \[1:3\].
Coordinates of P are ,
\[(\dfrac{{1( - 3) + 3(1)}}{{1 + 3}},\dfrac{{1(1) + 3(3)}}{{1 + 3}}) = (0,\dfrac{5}{2})\]
Direct common tangents are tangents drawn from the point Q which divides \[{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\]​ externally in the ratio \[{\text{1:3}}\]
Coordinates of point Q are,
\[(\dfrac{{1( - 3) - 3(1)}}{{1 - 3}},\dfrac{{1(1) - 3(3)}}{{1 - 3}}) = (3,4)\].
so the coordinates of Q are tangents through the point P \[(0,\dfrac{5}{2})\]
now, let the equation line passing through point p be,
\[
  y = mx + c \\
   \Rightarrow \dfrac{5}{2} = m(0) + c \\
   \Rightarrow c = \dfrac{5}{2} \\
   \Rightarrow y = mx + \dfrac{5}{2} \\
  \]
Now, applying the condition of tangency of the above obtained line to any one of the circles.
\[
  for,y = mx + \dfrac{5}{2} \\
  \therefore \left| {\dfrac{{m.1 - 3 + \dfrac{5}{2}}}{{\sqrt {{m^2} + 1} }}} \right| = 1 \\
  {(m - \dfrac{1}{2})^2} = ({m^2} + 1) \\
  m = \infty ,\dfrac{{ - 3}}{4} \\
  \]
So , the equation of line are
\[
  x = 0 \\
  y - \dfrac{5}{2} = \dfrac{{ - 3x}}{4} \\
  \]
Now, tangents that are drawn from the point Q
Proceeding in the same manner as that for transverse tangent
\[
  y = mx + c \\
   \Rightarrow 4 = 3m + c \\
   \Rightarrow c = 4 - 3m \\
   \Rightarrow y = mx + 4 - 3m \\
  \]
Applying the condition of tangency for any one of the circle
\[
   \Rightarrow \left| {\dfrac{{m.1 - 3 + 4 - 3m}}{{\sqrt {{m^2} + 1} }}} \right| = 1 \\
   \Rightarrow {( - 2m + 1)^2} = {m^2} + 1 \\
   \Rightarrow 4{m^2} - 4m + 1 = {m^2} + 1 \\
   \Rightarrow 3{m^2} - 4m = 0 \\
   \Rightarrow m = 0,\dfrac{4}{3} \\
  \]
Hence the equation of transverse tangents are
\[
  y = 4 \\
  4x - 3y = 0 \\
  \]
Hence, all the possible four tangent equations are stated as above.

Note: A tangent to a circle is defined as a line that passes through exactly one point on a circle, and is perpendicular to a line passing through the centre of the circle. A line that is tangent to more than one circle is referred to as a common tangent of both circles.