Area of a triangle formed by three tangent lines of three tangent circles

GarryMan

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I unfortunately made an error in my placement of the largest circle, so I am going to work the problem again, from start to finish, using the correct location.

Let's begin by placing the two smaller circles in the plane as follows:

View attachment 13003

Now, the center of the largest circle must satisfy the nonlinear system:

\(\displaystyle (x+5)^2+(y+5)^2=14^2\)

\(\displaystyle (x+5)^2+(y-7)^2=16^2\)

Solving this system, and taking the solution which places the largest circle to the right of the others, we obtain:

\(\displaystyle (x,y)=\left(\frac{7}{2}\sqrt{15}-5,-\frac{3}{2}\right)\)

And so we now place the largest circle in the plane:

View attachment 13004

The vertical tangent line (tangent to the largest circle) will be:

\(\displaystyle x=\frac{7}{2}\sqrt{15}-14\)

The line tangent to the circle of radius 7 will have a slope of:

\(\displaystyle m=\frac{-\dfrac{3}{2}+5}{\dfrac{7}{2}\sqrt{15}-5-(-5)}=\frac{1}{\sqrt{15}}\)

The tangent point is obtained through:

\(\displaystyle \frac{7-y_1}{-5-x_1}=-\sqrt{15}\)

\(\displaystyle (x_1+5)^2+(y_1-7)^2=7^2\)

This gives us the tangent point:

\(\displaystyle \left(x_1,y_1\right)=\left(-\frac{13}{4},\frac{7}{4}(4-\sqrt{15})\right)\)

And so this tangent line has the equation:

\(\displaystyle y=\frac{1}{\sqrt{15}}x+7-\frac{23}{\sqrt{15}}\)

For the line tangent to the smallest circle, it's slope is:

\(\displaystyle m=\frac{-\dfrac{3}{2}-7}{\dfrac{7}{2}\sqrt{15}-5-(-5)}=-\frac{17}{7\sqrt{15}}\)

The tangent point is obtained through:

\(\displaystyle \frac{-5-y_1}{-5-x_1}=\frac{7\sqrt{15}}{17}\)

\(\displaystyle (x_1+5)^2+(y_1+5)^2=5^2\)

This gives us the tangent point:

\(\displaystyle \left(x_1,y_1\right)=\left(-\frac{75}{32},\frac{35}{32}\sqrt{15}-5\right)\)

And so this tangent line has the equation:

\(\displaystyle y=-\frac{17}{7\sqrt{15}}x+\frac{817}{224}\sqrt{\frac{3}{5}}-5\)

Here is a plot including the three tangent lines:

View attachment 13005

We then find that the length of the vertical side of the triangle whose area we are finding is:

\(\displaystyle \left(\frac{21}{2}-\frac{37}{\sqrt{15}}\right)-\left(\frac{10067}{224\sqrt{15}}-\frac{27}{2}\right)=\frac{1}{672}\left(16128-3671\sqrt{15}\right)\)

The area of the triangle whose vertices are at the centers of the circles is:

\(\displaystyle A_V=\frac{1}{2}(7-(-5))\left(\frac{7}{2}\sqrt{15}-5-(-5)\right)=21\sqrt{15}\)

Also:

\(\displaystyle A_V=k(12)^2\implies k=\frac{7\sqrt{15}}{48}\)

Hence, the area we want is:

\(\displaystyle A=\frac{7\sqrt{15}}{48}\left(\frac{1}{672}\left(16128-3671\sqrt{15}\right)\right)^2=\frac{154085333\sqrt{15}-592058880}{1032192}\approx4.56412\)

I sure hope I didn't make any other errors!! :p
How exactly did you get the values that you are using to find the vertical side of the triangle? I am also working on this problem and struggling to understand what you did here.
 

MarkFL

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How exactly did you get the values that you are using to find the vertical side of the triangle? I am also working on this problem and struggling to understand what you did here.
If I understand you correctly, I evaluated the two non-vertical tangent lines at the \(x\)-value of the vertical tangent line.
 

GarryMan

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If I understand you correctly, I evaluated the two non-vertical tangent lines at the \(x\)-value of the vertical tangent line.
And what exactly did you mean when you stated that k(12)^2 near the very end of the problem
 

MarkFL

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And what exactly did you mean when you stated that k(12)^2 near the very end of the problem
The area of a plane figure will vary as the square of any of its linear measures. If we have two similar figures, then the constant of proportionality will be the same, when we use corresponding linear measures.
 
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