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

Last updated date: 13th Jun 2024
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Hint: We first find the centres A,B and their radii $r,s$ of the given circles from the standard equation in two variables. We compare AB and $r+s$ to find that there will be four tangents for the non-intersecting circles. We find the coordinate of the point of intersection of transverse tangent P and the direct tangents Q using section formula. We use the slope-point form of the equation of the line with slope $m$, point P, Q, and then the condition of tangency on circles to find $m.$ 

Complete step-by-step solution
We know from section formula that any point $R(x,y)$ which divides a line segment internally $\overline{AB}$ in a ratio$AP:PB=m:n$ with endpoints $A({{x}_{1}},{{y}_{2}})\text{ and B(}{{\text{x}}_{2}}\text{,}{{\text{y}}_{2}}\text{)}$ then the coordinates of R are
$R\left( x=\dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},y=\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)$
If $R$ divides the line segment $\overline{AB}$ externally, then the coordinates of R are
$R\left( x=\dfrac{m{{x}_{2}}-n{{x}_{1}}}{m+-n},y=\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n} \right)$
We also know tangency condition that the line $ax+by+c=0$ will be tangent to the circle with centre $\left( -g,-f \right)$ and radius $r$ if
$r=\dfrac{a\left( -g \right)+b\left( -f \right)+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
We know from the standard equation of circle in two variable form is
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
The centre of the above circle is given by the co-ordinates $\left( -g,-f \right)$ and radius is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. The given equation of circles is
\begin{align} & {{x}^{2}}+{{y}^{2}}-2x-6y+9=0....\left( 1 \right) \\ & {{x}^{2}}+{{y}^{2}}+6x-2y+1=0....\left( 2 \right) \\ \end{align}
We compare the equation of circle (1) and (2) with the standard equation find their respective centres $A,B$ and radii $r,s$. We have
\begin{align} & A\left( -\dfrac{\left( -2 \right)}{2},-\dfrac{\left( -6 \right)}{2} \right)=A\left( 1,3 \right) \\ & r=\sqrt{{{1}^{2}}+{{3}^{2}}-9}=1 \\ & B\left( -\dfrac{6}{2},-\dfrac{\left( -2 \right)}{2} \right)=B\left( -3,1 \right) \\ & s=\sqrt{9+1-1}=3 \\ \end{align}
Let us find the distance between two centres AB using two point distance formula. We have
$AB=\sqrt{{{\left( -3-1 \right)}^{2}}+{{\left( 3-\left( -1 \right) \right)}^{2}}}=\sqrt{20}$
The sum of the radii are $r+s=4$. So we have $AB>r+s$ which means that the circle do not intersect each other . So we will have four tangents called two transverse tangents and two direct tangents.
We know that the point of intersection of transverse tangents divides the line segment joining the centers internally at the ratio of their radii and the point of intersection of direct tangents divides the same line segment externally at the ratio of their radii. Let the point of intersection of transverse tangents be P and the point of intersection of direct tangents be Q. The ratio of the radii is $r:s=1:3$. 

We use the section formula for internal division and find the co-ordinate of P which divides the line segment joining $A\left( 1,3 \right),B\left( -3,1 \right)$at a ratio 1:3. We have
$P\left( \dfrac{1\left( -3 \right)+3\cdot 1}{1+3},\dfrac{1\cdot 1+3\cdot 3}{1+3} \right)=P\left( 0,\dfrac{5}{2} \right)$
We use the slope point form to find the equation of tangents passing thought $P\left( 0,\dfrac{5}{2} \right)$ as
\begin{align} & y-\dfrac{5}{2}=m\left( x-0 \right) \\ & \Rightarrow mx-y+\dfrac{5}{2}=0....(1) \\ \end{align}
We use the condition of tangency for above line for circle (1) with centre $\left( 1,3 \right)$ and radius 1. We have,
\begin{align} & 1=\dfrac{m\cdot 1-3+\dfrac{5}{2}}{\sqrt{{{m}^{2}}+1}} \\ & \Rightarrow {{\left( m-\dfrac{1}{2} \right)}^{2}}={{m}^{2}}+1 \\ & \Rightarrow {{m}^{2}}-m+\dfrac{1}{4}={{m}^{2}}+1 \\ & \Rightarrow m+\dfrac{3}{4}=0 \\ \end{align}
So the value of the slopes is $m=\dfrac{-3}{4}$ and $m=\infty$ as the coefficient ${{m}^{2}}$ is zero. We put these values in equation (1) and obtain the equation of transverse tangents as
\begin{align} & -3x-4y+10=0 \\ & x=0 \\ \end{align}
We use the section formula for external division and find the co-ordinate of Q which divides the line segment joining $A\left( 1,3 \right),B\left( -3,1 \right)$at a ratio 1:3. We have
$Q\left( \dfrac{1\left( -3 \right)-3\cdot 1}{1-3},\dfrac{1\cdot 1-3\cdot 3}{1-3} \right)=Q\left( 3,4 \right)$
We use the slope point form to find the equation of tangents passing thought $Q\left( 3,4 \right)$ as
\begin{align} & y-4=m\left( x-3 \right) \\ & \Rightarrow mx-y+4-3m=0....(2) \\ \end{align}
We use the condition of tangency for above line for circle (1) with centre $\left( 1,3 \right)$ and radius 1. We have,
\begin{align} & 1=\dfrac{m\cdot 1-3+4-3m}{\sqrt{{{m}^{2}}+1}} \\ & \Rightarrow {{\left( -2m-1 \right)}^{2}}={{m}^{2}}+1 \\ & \Rightarrow 4{{m}^{2}}+4m+1={{m}^{2}}+1 \\ & \Rightarrow m\left( 3m+4 \right)=0 \\ \end{align}
So we have the slopes of direct tangents as $m=0,\dfrac{-4}{3}$. We put the slopes in equation (2) and get the direct tangent equations as,
\begin{align} & y=4 \\ & 4x-3y=0 \\ \end{align}

Note: We can alternatively use tangency condition on circle (2). There will be three common tangents to both the circles when they touch externally $AB=r+s$, two common tangents if they intersect $\left| r-s \right| < AB < r+s$,one common tangent when they touch internally $\left | r-s \right|=AB$ and no common tangent when one lie in the interior of other $\left| r-s \right| > AB.$