Question

Find the common tangents equation of the circles ${{x}^{2}}+{{y}^{2}}={{c}^{2}},{{\left( x-a \right)}^{2}}+{{y}^{2}}={{b}^{2}}$. \[\]

Answer 1

Hint: We compare the given equation of circles with standard equation of circle in centre –radius form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ to find the centres of the radii of both circles. We use the equation of any tangent of any circle with radius $r$ that is $y=mx+c\sqrt{1+{{m}^{2}}}$ to find the tangent of the equation of the first circle. We equate the radius of the second circle with distance of the centre of second circle to the tangent of the first circle to find different values of $m$ in terms of $a,b,c$ ; for each value of we shall get one tangent equation.\[\]

Complete step by step answer:

We are given the following equations of circle

\[\begin{align}

  & {{x}^{2}}+{{y}^{2}}={{c}^{2}}\Rightarrow {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{c}^{2}}......\left( 1 \right) \\

 & {{\left( x-a \right)}^{2}}+{{y}^{2}}={{b}^{2}}\Rightarrow {{\left( x-a \right)}^{2}}+\left( y-0 \right)={{b}^{2}}.....\left( 2 \right) \\

\end{align}\]

We see that the coordinate of the centre of the first circle is $\left( 0,0 \right)$ and radius is $c$. The equation of tangent of the circle (1) is given by

\[\begin{align}

  & y=mx+c\sqrt{1+{{m}^{2}}} \\

 & \Rightarrow mx-y+c\sqrt{1+{{m}^{2}}}=0....\left( 3 \right) \\

\end{align}\]

We see that the coordinate of the centre of the second circle is $\left( a,0 \right)$ and radius is $b$. Let tangent (3) be the common tangent of circle (1) and circle (2). So the radius of circle (2) is equal to the distance from the centre $\left( a,0 \right)$ to the tangent line $mx-y+c\sqrt{1+{{m}^{2}}}=0$. So we have

\[\begin{align}

  & b=\dfrac{\left| am-1\cdot 0+c\sqrt{1+{{m}^{2}}} \right|}{\sqrt{1+{{m}^{2}}}} \\

 & \Rightarrow \pm b=\dfrac{am+c\sqrt{1+{{m}^{2}}}}{\sqrt{1+{{m}^{2}}}} \\

 & \Rightarrow \pm b\sqrt{1+{{m}^{2}}}+c\sqrt{1+{{m}^{2}}}=am. \\

 & \Rightarrow \left( \pm b-c \right)\sqrt{1+{{m}^{2}}}=am \\

\end{align}\]

We take positive sign for $b$ in the above equation and the square both sides to have;

\[\begin{align}

  & {{\left( b-c \right)}^{2}}\left( 1-{{m}^{2}} \right)={{a}^{2}}{{m}^{2}} \\

 & \Rightarrow {{m}^{2}}\left( {{a}^{2}}-{{\left( b-c \right)}^{2}} \right)={{\left( b-c \right)}^{2}} \\

 & \Rightarrow {{m}^{2}}=\dfrac{{{\left( b-c \right)}^{2}}}{{{a}^{2}}-{{\left( b-c \right)}^{2}}} \\

\end{align}\]

So the roots of the equation are

\[m=\dfrac{\pm \left( b-c \right)}{\sqrt{{{a}^{2}}-{{\left( b-c \right)}^{2}}}}\]

We take positive sign for $b$ in the above equation (4) and the square both sides to have;

\[\begin{align}

  & {{\left( -b-c \right)}^{2}}\left( 1-{{m}^{2}} \right)={{a}^{2}}{{m}^{2}} \\

 & \Rightarrow {{\left( b+c \right)}^{2}}\left( 1-{{m}^{2}} \right)={{a}^{2}}{{m}^{2}} \\

 & \Rightarrow {{m}^{2}}\left( {{a}^{2}}-{{\left( b+c \right)}^{2}} \right)={{\left( b+c \right)}^{2}} \\

 & \Rightarrow {{m}^{2}}=\dfrac{{{\left( b+c \right)}^{2}}}{{{a}^{2}}-{{\left( b+c \right)}^{2}}} \\

\end{align}\]

So the roots of the equation are

\[m=\dfrac{\pm \left( b+c \right)}{\sqrt{{{a}^{2}}-{{\left( b+c \right)}^{2}}}}\]

So the equations of tangents are

\[mx-y+c\sqrt{1+{{m}^{2}}}=0\text{ where }m=\dfrac{\pm \left( b-c \right)}{\sqrt{{{a}^{2}}-{{\left( b-c \right)}^{2}}}},\dfrac{\pm \left( b+c \right)}{\sqrt{{{a}^{2}}-{{\left( b+c \right)}^{2}}}}\]

Note: We note that the perpendicular distance between a point $\left( {{x}_{1}},{{y}_{1}} \right)$ from a line $ax+by+c=0$ is given by $d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ . The number of tangents will depend upon the values of $a,b,c$. If the sum of the radii $b+c$ is less than distance between the centres of the circles $a$ then we shall have four common tangents for no-intersecting external circles, if $b+c=a$ then we have 3 common tangents for circles touching externally , if difference between radii $\left| b-c \right| < a < b+c$ then we have 2 common tangents from intersecting circles and if \[a=\left| b-c \right|\]the 1 common tangent circles touching internally.