Question

How do you find the equations of common tangents to the circles
 \[{{x}^{2}}+{{y}^{2}}=9,{{x}^{2}}+{{y}^{2}}-16x+2y+49=0\] ?

Answer 1

Hint: Firstly , the radii for the circles are found then, the equations of the tangents of a circle are found by considering the equation for slope, And further solving the equations by finding the values of \[m\] and the different cases are taken then, the equations of tangents that are converse to the circle will be found.
Radius of the circle with origin as center
 \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] where \[a\] is radius
And with center \[\left( h,k \right)\] is
 \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]
Where \[r\] is radius

Complete step-by-step answer:
The circles are
 \[
   {{x}^{2}}+{{y}^{2}}=9 \\
  {{x}^{2}}+{{y}^{2}}-16x+2y+49=0 \;
\]
And the centers are \[A\left( 0,0 \right)\] , \[B\left( 8,-1 \right)\]
Here, as we can see that the radii for circles are
 \[
   r_1=3 \\
  r_2=\sqrt{64+1-49}=\sqrt{16}=4 \;
\]
Here, if we calculate the length of \[AB\]
 \[
   AB=\sqrt{{{\left( 0-8 \right)}^{2}}+{{\left( 0+1 \right)}^{2}}} \\
  \Rightarrow AB=\sqrt{65}>r_1+r_2 \;
\]
Here, we see that the circles lie outside each other,
The external center of similitude S divides \[AB\] externally in the ratio is \[3:4\]
So, the coordinates are \[\left( -24,+3 \right)\]
Suppose \[m\] is the slope of the direct common tangents
 \[
   y-3=m\left( x+24 \right) \\
  \Rightarrow y-3=mx+24m \\
  \Rightarrow \left( mx-y \right)+\left( 24m+3 \right)=0---\left( 1 \right) \;
\]
This is a tangent to the circle \[{{x}^{2}}+{{y}^{2}}=9\]
 \[
   9\left( {{m}^{2}}+1 \right)=9{{\left( 8m+1 \right)}^{2}} \\
  \Rightarrow 9\left( {{m}^{2}}+1 \right)=64{{m}^{2}}+10m+1 \\
  63{{m}^{2}}+16m=0 \\
  m\left( 63m+16 \right)=0 \\
  \Rightarrow m=0or-\dfrac{16}{63} \;
\]
Case 1: If we take \[m=0\] ,
Substituting in the above equations, equation of tangent is
 \[
   -y+3=0 \\
  y-3=0 \;
\]
Case 2: If we take \[m=-\dfrac{16}{63}\]
Equation of the tangent is
 \[
   \Rightarrow -\dfrac{16}{63}x-y+\left( \dfrac{-384}{63}+3 \right)=0 \\
  \Rightarrow -\dfrac{16}{63}x-y+\dfrac{195}{63}=0 \\
  \Rightarrow 16x+63y+195=0 \;
\]
The internal center of similitude \[S'\] is dividing \[AB\] internally in the ratio of \[3:4\]
The coordinates of \[S'\] are found to be as \[\left( \dfrac{24}{7},-\dfrac{3}{7} \right)\]
The equation of the transverse common tangent is found as:
 \[
   \left( y+\dfrac{3}{7} \right)=m\left( x-\dfrac{24}{7} \right) \\
  \Rightarrow \dfrac{7y+3}{7}=m\left( \dfrac{7x-24}{7} \right) \\
  \Rightarrow 7y+3=7mx-24m \\
  \Rightarrow 7mx-7y-\left( 24m+3 \right)=0---\left( 2 \right) \;
\]
This is a tangent to the circle \[{{x}^{2}}+{{y}^{2}}=9\]
 \[3=\dfrac{\left| 24m+3 \right|}{\sqrt{49{{m}^{2}}+49}}=\dfrac{3}{7}\dfrac{\left| 28m+1 \right|}{\sqrt{{{m}^{2}}+1}}\]
Solving this, we get
 \[
   49\left( {{m}^{2}}+1 \right)={{\left( 8m+1 \right)}^{2}} \\
  \Rightarrow 49{{m}^{2}}+49=64{{m}^{2}}+16m+1 \\
  \Rightarrow 15{{m}^{2}}+16m-48=0 \\
  \Rightarrow \left( 3m-4 \right)\left( 5m+12 \right)=0 \\
  \Rightarrow m=\dfrac{4}{3}or-\dfrac{12}{5} \\
\]
Case (i), the equation of tangent is
 \[
   \dfrac{28}{x}\centerdot x-7y-\left( \dfrac{96}{3}+3 \right)=0 \\
  \Rightarrow \dfrac{28}{x}\centerdot x-7y-\dfrac{105}{3}=0 \\
  \Rightarrow \dfrac{7}{3}\left( 4x-3y-15 \right)=0 \\
  \Rightarrow 4x-3y-15=0 \;
\]
Taking Case (ii)
 \[m=-\dfrac{12}{5}\]
Equation of the transverse common tangent is
 \[
   -\dfrac{84}{5}x-7y-\left( -\dfrac{288}{5}+3 \right)=0 \\
  \Rightarrow -\dfrac{84}{5}x-7y+\dfrac{273}{5}=0 \\
  \Rightarrow -\dfrac{7}{5}\left( 12x+5y-39 \right)=0 \\
  \Rightarrow 12x+5y-39=0 \\
\]
Therefore, equation of direct common tangents are
 \[
   y-3=0 \\
  16x+63y+195=0 \;
\]
Hence, equation of transverse common tangent are
 \[4x-3y-15=0,12x+5y-39=0\]
So, the correct answer is “ \[4x-3y-15=0,12x+5y-39=0\] ”.

Note: The equations of the tangents of a circle are found by considering the equation for slope,
 \[y-3=m\left( x+24 \right)\]
And further solving the equations by finding the values of \[m\] and the different cases are taken then, the equations of tangents that are converse to the circle are also found.