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.