Find the length of common chord of the circles ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ and ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$. Find also the equation of common tangents and show that the length of each is $4a;a >0$ .
Answer
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Hint: We start solving the problem by finding the center and radius of both circles. We now find the length of the common chord by finding the equation and perpendicular distance from the center to the chord. We use the slope equation form for the tangent and find the slope using the radius of another circle. We now draw a parallel line segment to the tangent from the center of the circle to find the length of the tangent.
Complete step by step solution:
Given that we have two circles whose equations are given by ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ and ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$.
We need to find the length of the common chord, the equation of common tangent, and the length of each common tangent.
Let us first find the center and radius of the circles.
For the circle, ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$
We can rewrite this circle equation as,
$\Rightarrow {{x}^{2}}-2ax+{{a}^{2}}+{{y}^{2}}-4ay+4{{a}^{2}}-{{a}^{2}}-4{{a}^{2}}-4{{a}^{2}}=0$
Now upon grouping the terms we get,
${{\left( x-a \right)}^{2}}+{{\left( y-2a \right)}^{2}}-9{{a}^{2}}=0$
${{\left( x-a \right)}^{2}}+{{\left( y-2a \right)}^{2}}={{\left( 3a \right)}^{2}}$
We know that from the equation of the circle ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$, we get center as $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius as r.
Here, ${{x}_{1}}=a;{{y}_{1}}=2a;r=3a$
Hence the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ as ${{C}_{1}}\left( a,2a \right)$ and ${{r}_{1}}=3a$.
Now let us find the center and radius of the circle, ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$
We can rewrite this circle equation as,
$\Rightarrow {{x}^{2}}-3ax+\dfrac{9{{a}^{2}}}{4}+{{y}^{2}}+4ay+4{{a}^{2}}-\dfrac{9{{a}^{2}}}{4}-4{{a}^{2}}=0$
Now upon grouping the terms we get,
${{\left( x-\dfrac{3a}{2} \right)}^{2}}+{{\left( y+2a \right)}^{2}}-\dfrac{25{{a}^{2}}}{4}=0$
$\Rightarrow {{\left( x-\dfrac{3a}{2} \right)}^{2}}+{{\left( y+2a \right)}^{2}}={{\left( \dfrac{5a}{2} \right)}^{2}}$
We know that from the equation of the circle ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$, we get center as $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius as r.
Here, ${{x}_{1}}=\dfrac{3a}{2};{{y}_{1}}=-2a;r=\dfrac{5a}{2}$
Hence, the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$ as ${{C}_{2}}\left( \dfrac{3a}{2},-2a \right)$ and ${{r}_{2}}=\dfrac{5a}{2}$.
Now to find the length of the common chord,
First, let us find the equation of the common chord.
It is given by ${{s}_{1}}-{{s}_{2}}=0$ , given that ${{s}_{1}},{{s}_{2}}$ are circle equations.
Hence upon substituting the circle equations we get, $\Rightarrow {{x}^{2}}+{{y}^{2}}-3ax+4ay-\left( {{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}} \right)=0$$\Rightarrow {{x}^{2}}+{{y}^{2}}-3ax+4ay-{{x}^{2}}-{{y}^{2}}+2ax+4ay+4{{a}^{2}}=0$
Upon evaluating further, we get,
$\Rightarrow -ax+8ay+4{{a}^{2}}=0$.
We can take $a$ common.
$\Rightarrow -a(x-8y-4a=0)$
Hence, the equation of the common chord is $x-8y-4a=0$
Now to find the length EK, we use the Pythagoras theorem for the right triangle ${{C}_{1}}KE$ formed.
The formula will hence be, $\sqrt{{{r}^{2}}-{{d}^{2}}}$
Here, $d$ is the distance between the center of any circle to the chord and $r$ is the radius of that circle.
Let us consider the circle ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ for calculating the common chord.
The distance from the center $\left( a,2a \right)$ to the line $x-8y-4a=0$ can be calculated by,
The perpendicular distance from the point $\left( {{x}_{1}},{{y}_{1}} \right)$ to the line $ax+by+c=0$ is $\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Upon substituting the values, we get,
$\Rightarrow d=\dfrac{\left| a-8\left( 2a \right)-4a \right|}{\sqrt{{{1}^{2}}+{{\left( -8 \right)}^{2}}}}$.
$\Rightarrow d=\dfrac{\left| -16a-3a \right|}{\sqrt{1+64}}$.
On further evaluation we get,
$\Rightarrow d=\dfrac{19a}{\sqrt{65}}$
Now we know that radius is $3a\;$
Now let us substitute in the length of the common chord formula.
The length is given by, $\sqrt{{{r}^{2}}-{{d}^{2}}}$
$\Rightarrow \sqrt{{{(3a)}^{2}}-{{\left( \dfrac{19a}{\sqrt{65}} \right)}^{2}}}$
$\Rightarrow \sqrt{9{{a}^{2}}-\left( \dfrac{361{{a}^{2}}}{65} \right)}$
On further evaluating we get,
$\Rightarrow \sqrt{\left( \dfrac{585{{a}^{2}}-361{{a}^{2}}}{65} \right)}$
$\Rightarrow \sqrt{\left( \dfrac{224{{a}^{2}}}{65} \right)}$
On further simplification,
$\Rightarrow \sqrt{\left( \dfrac{16\times 14}{65} \right)}\times a$
$\Rightarrow \dfrac{4\sqrt{14}a}{\sqrt{65}}$
From the figure, we can see that the length which we just found out is EK and the length of the common chord is twice the length of EK
The length of the common chord will now be,
$\Rightarrow 2\times \dfrac{4\sqrt{14}a}{\sqrt{65}}=\dfrac{8\sqrt{14}a}{\sqrt{65}}$
Hence, the length of the common chord is $\dfrac{8\sqrt{14}a}{\sqrt{65}}$
Now we are also asked to find the equations for the common tangents.
Let us understand this diagrammatically.
Now, draw the common tangents for the circles. We must find the equations of the line segment of common tangent GH and IJ.
We know that the general equation of the tangent of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\;$ with a slope ‘m’ is given as $y+f=m\left( x+g \right)\pm r\left( \sqrt{1+{{m}^{2}}} \right)$, where ‘r’ is the radius of the circle.
So, the general equation of the tangent of the circle ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ is $\Rightarrow y-2a=m\left( x-a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right)$
The slope equation of the common tangent is given by,
$\Rightarrow m\left( x-a \right)-\left( y-2a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right)=0$.
This tangent is also tangent for the circle ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$.
We know that the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle and we know that the center of the circle is ${{C}_{2}}\left( \dfrac{3a}{2},-2a \right)$ and radius is ${{r}_{2}}=\dfrac{5a}{2}$
So, from the above diagram, we know $H{{C}_{2}}={{r}_{2}}$
By using the perpendicular distance formula, we can find $H{{C}_{2}}$
Hence,
$\Rightarrow \dfrac{\left| m\left( \dfrac{3a}{2}-a \right)-\left( -2a-2a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right) \right|}{\sqrt{{{m}^{2}}+1}}=\dfrac{5a}{2}$
$\Rightarrow \left| m\left( \dfrac{a}{2} \right)-\left( -4a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right) \right|=\dfrac{5a}{2}\sqrt{{{m}^{2}}+1}$
$\Rightarrow \left| \dfrac{ma}{2}+4a\pm 3a\left( \sqrt{1+{{m}^{2}}} \right) \right|=\dfrac{5a}{2}\sqrt{{{m}^{2}}+1}$
Upon further evaluation we get,
$\Rightarrow \dfrac{\left| ma+8a\pm 6a\left( \sqrt{1+{{m}^{2}}} \right) \right|}{2}=\dfrac{5a}{2}\sqrt{{{m}^{2}}+1}$
$\Rightarrow \left| ma+8a\pm 6a\left( \sqrt{1+{{m}^{2}}} \right) \right|=5a\sqrt{{{m}^{2}}+1}$
$\Rightarrow ma+8a\pm 6a\left( \sqrt{1+{{m}^{2}}} \right)=\pm 5a\sqrt{{{m}^{2}}+1}$
Now, opening the $\pm$ sign,
We get two equations,
$\Rightarrow ma+8a+6a\left( \sqrt{1+{{m}^{2}}} \right)=5a\sqrt{{{m}^{2}}+1}$ and $ma+8a-6a\left( \sqrt{1+{{m}^{2}}} \right)=5a\sqrt{{{m}^{2}}+1}$
$\Rightarrow ma+8a=-a\sqrt{{{m}^{2}}+1}$ and $ma+8a=11a\sqrt{{{m}^{2}}+1}$
$\Rightarrow m+8=-\sqrt{{{m}^{2}}+1}$ and $m+8=11\sqrt{{{m}^{2}}+1}$
$\Rightarrow {{m}^{2}}+64+16m={{m}^{2}}+1$ and ${{m}^{2}}+16m+64=121\left( {{m}^{2}}+1 \right)$
We know to get,
$\Rightarrow 63+16m=0$ and $120{{m}^{2}}-16m+57=0$.
Since $16m=-63$ and the other quadratic equation is not solvable,
We can conclude that $m=\dfrac{-63}{16}$
Upon substituting this value in $y-2a=m\left( x-a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right)$
Hence, the equations of the common tangents are given by,
$\Rightarrow \dfrac{-63}{16}\left( x-a \right)-\left( y-2a \right)\pm 3a\left( \sqrt{1+{{\left( \dfrac{-63}{16} \right)}^{2}}} \right)=0$
Now, we are asked to find the lengths of the line segment of common tangent $GH$ and $IJ$ . Let us draw a line segment parallel to the tangent $GH$ . Here we can see that the length of $HK$ is equal to the length of $G{{C}_{1}}$. Since $HK$ is perpendicular to $GH$ and ${{C}_{1}}K$ is parallel to $GH$ , and the value of the angle $K$ is ${{90}^{o}}$
We know the value of the length of $K{{C}_{2}}={{r}_{2}}-{{r}_{1}}$ as $H{{C}_{2}}$ is the radius of the circle.
Now that we have the triangle ${{C}_{1}}K{{C}_{2}}$ as a right-angled triangle,
We use the Pythagoras theorem.
$\Rightarrow {{C}_{1}}{{C}_{2}}^{2}=K{{C}_{2}}^{2}+{{C}_{1}}{{K}^{2}}$
Let us find ${{C}_{1}}{{C}_{2}}$ first.
Now we find the distance between the centers of the two circles. We know that distance between points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{{{\left( \dfrac{3a}{2}-a \right)}^{2}}+{{\left( -2a-2a \right)}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{{{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( -4a \right)}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{\dfrac{{{a}^{2}}}{4}+16{{a}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\dfrac{\sqrt{65}a}{2}$
Now on substituting in the Pythagoras formula,
$\Rightarrow {{\left( \dfrac{\sqrt{65}a}{2} \right)}^{2}}={{\left( {{r}_{2}}-{{r}_{1}} \right)}^{2}}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{65{{a}^{2}}}{4}={{\left( \dfrac{5a}{2}-3a \right)}^{2}}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{65{{a}^{2}}}{4}={{\left( \dfrac{-a}{2} \right)}^{2}}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{65{{a}^{2}}}{4}=\dfrac{{{a}^{2}}}{4}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{64{{a}^{2}}}{4}={{C}_{1}}{{K}^{2}}$
$\Rightarrow 16{{a}^{2}}={{C}_{1}}{{K}^{2}}$
$\Rightarrow 4a={{C}_{1}}K$
Since ${{C}_{1}}K$ is equal to the length of the common tangents we get the length of the common tangents as $4a;a>0$
Hence, the length of the common chord is $d=\dfrac{8\sqrt{14}a}{\sqrt{65}}$ , equations of the common tangents are $\dfrac{-63}{16}\left( x-a \right)-\left( y-2a \right)\pm 3a\left( \sqrt{1+{{\left( \dfrac{-63}{16} \right)}^{2}}} \right)=0$ and we have proved the length of the common tangents as $4a;a>0$
Note: We should add that the distance between the centers is equal to the length of the tangent as it is not correct for intersecting circles. We should know that the line connecting the center and the point of contact makes a right angle with the tangent. Chords only occur in intersecting circles. In externally touching circles we get this chord as a tangent.
Complete step by step solution:
Given that we have two circles whose equations are given by ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ and ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$.
We need to find the length of the common chord, the equation of common tangent, and the length of each common tangent.
Let us first find the center and radius of the circles.
For the circle, ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$
We can rewrite this circle equation as,
$\Rightarrow {{x}^{2}}-2ax+{{a}^{2}}+{{y}^{2}}-4ay+4{{a}^{2}}-{{a}^{2}}-4{{a}^{2}}-4{{a}^{2}}=0$
Now upon grouping the terms we get,
${{\left( x-a \right)}^{2}}+{{\left( y-2a \right)}^{2}}-9{{a}^{2}}=0$
${{\left( x-a \right)}^{2}}+{{\left( y-2a \right)}^{2}}={{\left( 3a \right)}^{2}}$
We know that from the equation of the circle ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$, we get center as $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius as r.
Here, ${{x}_{1}}=a;{{y}_{1}}=2a;r=3a$
Hence the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ as ${{C}_{1}}\left( a,2a \right)$ and ${{r}_{1}}=3a$.
Now let us find the center and radius of the circle, ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$
We can rewrite this circle equation as,
$\Rightarrow {{x}^{2}}-3ax+\dfrac{9{{a}^{2}}}{4}+{{y}^{2}}+4ay+4{{a}^{2}}-\dfrac{9{{a}^{2}}}{4}-4{{a}^{2}}=0$
Now upon grouping the terms we get,
${{\left( x-\dfrac{3a}{2} \right)}^{2}}+{{\left( y+2a \right)}^{2}}-\dfrac{25{{a}^{2}}}{4}=0$
$\Rightarrow {{\left( x-\dfrac{3a}{2} \right)}^{2}}+{{\left( y+2a \right)}^{2}}={{\left( \dfrac{5a}{2} \right)}^{2}}$
We know that from the equation of the circle ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$, we get center as $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius as r.
Here, ${{x}_{1}}=\dfrac{3a}{2};{{y}_{1}}=-2a;r=\dfrac{5a}{2}$
Hence, the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$ as ${{C}_{2}}\left( \dfrac{3a}{2},-2a \right)$ and ${{r}_{2}}=\dfrac{5a}{2}$.
Now to find the length of the common chord,
First, let us find the equation of the common chord.
It is given by ${{s}_{1}}-{{s}_{2}}=0$ , given that ${{s}_{1}},{{s}_{2}}$ are circle equations.
Hence upon substituting the circle equations we get, $\Rightarrow {{x}^{2}}+{{y}^{2}}-3ax+4ay-\left( {{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}} \right)=0$$\Rightarrow {{x}^{2}}+{{y}^{2}}-3ax+4ay-{{x}^{2}}-{{y}^{2}}+2ax+4ay+4{{a}^{2}}=0$
Upon evaluating further, we get,
$\Rightarrow -ax+8ay+4{{a}^{2}}=0$.
We can take $a$ common.
$\Rightarrow -a(x-8y-4a=0)$
Hence, the equation of the common chord is $x-8y-4a=0$
Now to find the length EK, we use the Pythagoras theorem for the right triangle ${{C}_{1}}KE$ formed.
The formula will hence be, $\sqrt{{{r}^{2}}-{{d}^{2}}}$
Here, $d$ is the distance between the center of any circle to the chord and $r$ is the radius of that circle.
Let us consider the circle ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ for calculating the common chord.
The distance from the center $\left( a,2a \right)$ to the line $x-8y-4a=0$ can be calculated by,
The perpendicular distance from the point $\left( {{x}_{1}},{{y}_{1}} \right)$ to the line $ax+by+c=0$ is $\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Upon substituting the values, we get,
$\Rightarrow d=\dfrac{\left| a-8\left( 2a \right)-4a \right|}{\sqrt{{{1}^{2}}+{{\left( -8 \right)}^{2}}}}$.
$\Rightarrow d=\dfrac{\left| -16a-3a \right|}{\sqrt{1+64}}$.
On further evaluation we get,
$\Rightarrow d=\dfrac{19a}{\sqrt{65}}$
Now we know that radius is $3a\;$
Now let us substitute in the length of the common chord formula.
The length is given by, $\sqrt{{{r}^{2}}-{{d}^{2}}}$
$\Rightarrow \sqrt{{{(3a)}^{2}}-{{\left( \dfrac{19a}{\sqrt{65}} \right)}^{2}}}$
$\Rightarrow \sqrt{9{{a}^{2}}-\left( \dfrac{361{{a}^{2}}}{65} \right)}$
On further evaluating we get,
$\Rightarrow \sqrt{\left( \dfrac{585{{a}^{2}}-361{{a}^{2}}}{65} \right)}$
$\Rightarrow \sqrt{\left( \dfrac{224{{a}^{2}}}{65} \right)}$
On further simplification,
$\Rightarrow \sqrt{\left( \dfrac{16\times 14}{65} \right)}\times a$
$\Rightarrow \dfrac{4\sqrt{14}a}{\sqrt{65}}$
From the figure, we can see that the length which we just found out is EK and the length of the common chord is twice the length of EK
The length of the common chord will now be,
$\Rightarrow 2\times \dfrac{4\sqrt{14}a}{\sqrt{65}}=\dfrac{8\sqrt{14}a}{\sqrt{65}}$
Hence, the length of the common chord is $\dfrac{8\sqrt{14}a}{\sqrt{65}}$
Now we are also asked to find the equations for the common tangents.
Let us understand this diagrammatically.
Now, draw the common tangents for the circles. We must find the equations of the line segment of common tangent GH and IJ.
We know that the general equation of the tangent of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\;$ with a slope ‘m’ is given as $y+f=m\left( x+g \right)\pm r\left( \sqrt{1+{{m}^{2}}} \right)$, where ‘r’ is the radius of the circle.
So, the general equation of the tangent of the circle ${{x}^{2}}+{{y}^{2}}-2ax-4ay-4{{a}^{2}}=0$ is $\Rightarrow y-2a=m\left( x-a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right)$
The slope equation of the common tangent is given by,
$\Rightarrow m\left( x-a \right)-\left( y-2a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right)=0$.
This tangent is also tangent for the circle ${{x}^{2}}+{{y}^{2}}-3ax+4ay=0$.
We know that the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle and we know that the center of the circle is ${{C}_{2}}\left( \dfrac{3a}{2},-2a \right)$ and radius is ${{r}_{2}}=\dfrac{5a}{2}$
So, from the above diagram, we know $H{{C}_{2}}={{r}_{2}}$
By using the perpendicular distance formula, we can find $H{{C}_{2}}$
Hence,
$\Rightarrow \dfrac{\left| m\left( \dfrac{3a}{2}-a \right)-\left( -2a-2a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right) \right|}{\sqrt{{{m}^{2}}+1}}=\dfrac{5a}{2}$
$\Rightarrow \left| m\left( \dfrac{a}{2} \right)-\left( -4a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right) \right|=\dfrac{5a}{2}\sqrt{{{m}^{2}}+1}$
$\Rightarrow \left| \dfrac{ma}{2}+4a\pm 3a\left( \sqrt{1+{{m}^{2}}} \right) \right|=\dfrac{5a}{2}\sqrt{{{m}^{2}}+1}$
Upon further evaluation we get,
$\Rightarrow \dfrac{\left| ma+8a\pm 6a\left( \sqrt{1+{{m}^{2}}} \right) \right|}{2}=\dfrac{5a}{2}\sqrt{{{m}^{2}}+1}$
$\Rightarrow \left| ma+8a\pm 6a\left( \sqrt{1+{{m}^{2}}} \right) \right|=5a\sqrt{{{m}^{2}}+1}$
$\Rightarrow ma+8a\pm 6a\left( \sqrt{1+{{m}^{2}}} \right)=\pm 5a\sqrt{{{m}^{2}}+1}$
Now, opening the $\pm$ sign,
We get two equations,
$\Rightarrow ma+8a+6a\left( \sqrt{1+{{m}^{2}}} \right)=5a\sqrt{{{m}^{2}}+1}$ and $ma+8a-6a\left( \sqrt{1+{{m}^{2}}} \right)=5a\sqrt{{{m}^{2}}+1}$
$\Rightarrow ma+8a=-a\sqrt{{{m}^{2}}+1}$ and $ma+8a=11a\sqrt{{{m}^{2}}+1}$
$\Rightarrow m+8=-\sqrt{{{m}^{2}}+1}$ and $m+8=11\sqrt{{{m}^{2}}+1}$
$\Rightarrow {{m}^{2}}+64+16m={{m}^{2}}+1$ and ${{m}^{2}}+16m+64=121\left( {{m}^{2}}+1 \right)$
We know to get,
$\Rightarrow 63+16m=0$ and $120{{m}^{2}}-16m+57=0$.
Since $16m=-63$ and the other quadratic equation is not solvable,
We can conclude that $m=\dfrac{-63}{16}$
Upon substituting this value in $y-2a=m\left( x-a \right)\pm 3a\left( \sqrt{1+{{m}^{2}}} \right)$
Hence, the equations of the common tangents are given by,
$\Rightarrow \dfrac{-63}{16}\left( x-a \right)-\left( y-2a \right)\pm 3a\left( \sqrt{1+{{\left( \dfrac{-63}{16} \right)}^{2}}} \right)=0$
Now, we are asked to find the lengths of the line segment of common tangent $GH$ and $IJ$ . Let us draw a line segment parallel to the tangent $GH$ . Here we can see that the length of $HK$ is equal to the length of $G{{C}_{1}}$. Since $HK$ is perpendicular to $GH$ and ${{C}_{1}}K$ is parallel to $GH$ , and the value of the angle $K$ is ${{90}^{o}}$
We know the value of the length of $K{{C}_{2}}={{r}_{2}}-{{r}_{1}}$ as $H{{C}_{2}}$ is the radius of the circle.
Now that we have the triangle ${{C}_{1}}K{{C}_{2}}$ as a right-angled triangle,
We use the Pythagoras theorem.
$\Rightarrow {{C}_{1}}{{C}_{2}}^{2}=K{{C}_{2}}^{2}+{{C}_{1}}{{K}^{2}}$
Let us find ${{C}_{1}}{{C}_{2}}$ first.
Now we find the distance between the centers of the two circles. We know that distance between points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{{{\left( \dfrac{3a}{2}-a \right)}^{2}}+{{\left( -2a-2a \right)}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{{{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( -4a \right)}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{\dfrac{{{a}^{2}}}{4}+16{{a}^{2}}}$
$\Rightarrow {{C}_{1}}{{C}_{2}}=\dfrac{\sqrt{65}a}{2}$
Now on substituting in the Pythagoras formula,
$\Rightarrow {{\left( \dfrac{\sqrt{65}a}{2} \right)}^{2}}={{\left( {{r}_{2}}-{{r}_{1}} \right)}^{2}}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{65{{a}^{2}}}{4}={{\left( \dfrac{5a}{2}-3a \right)}^{2}}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{65{{a}^{2}}}{4}={{\left( \dfrac{-a}{2} \right)}^{2}}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{65{{a}^{2}}}{4}=\dfrac{{{a}^{2}}}{4}+{{C}_{1}}{{K}^{2}}$
$\Rightarrow \dfrac{64{{a}^{2}}}{4}={{C}_{1}}{{K}^{2}}$
$\Rightarrow 16{{a}^{2}}={{C}_{1}}{{K}^{2}}$
$\Rightarrow 4a={{C}_{1}}K$
Since ${{C}_{1}}K$ is equal to the length of the common tangents we get the length of the common tangents as $4a;a>0$
Hence, the length of the common chord is $d=\dfrac{8\sqrt{14}a}{\sqrt{65}}$ , equations of the common tangents are $\dfrac{-63}{16}\left( x-a \right)-\left( y-2a \right)\pm 3a\left( \sqrt{1+{{\left( \dfrac{-63}{16} \right)}^{2}}} \right)=0$ and we have proved the length of the common tangents as $4a;a>0$
Note: We should add that the distance between the centers is equal to the length of the tangent as it is not correct for intersecting circles. We should know that the line connecting the center and the point of contact makes a right angle with the tangent. Chords only occur in intersecting circles. In externally touching circles we get this chord as a tangent.
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