TheXSquaredFactor

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Nombre de usuarioTheXSquaredFactor
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 #1
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I think that this question is fairly difficult. I have added a few extra lines and points to the diagram where I saw fit:

Here are my initial observations in regard to this current diagram:

  • The x-axis, which is blue, is tangent to both circles.
  • The green line is also tangent to both circles.
  • The green and blue lines intersect at one common point on the x-axis. 

These initial observations are enough to get me started with this problem. I constructed a parallel line to the x-axis, \(\overleftrightarrow{AK}\)\(\angle BKA\) is, therefore, a right angle. Point K is located at (8,1) because it acquires the y-coordinate of Point A and the x-coordinate of Point B. \(\triangle BKA\) is a right triangle, and it is possible to find \(\tan(m\angle KAB)\)

 

\(\tan m\angle KAB=\frac{9-1}{8-2}=\frac{8}{6}=\frac{4}{3}\). Notice how 4/3 is also the slope of  \(\overleftrightarrow{AB}\). The actual measure of the angle is not really important.

 

Because \(\overleftrightarrow{AK}||\overleftrightarrow{JH}\)\(\angle HJB\cong\angle KAB\). By extension, \(m\angle HJB=m\angle KAB\). This also means that \(\tan m\angle HJB=\tan m\angle KAB=\frac{4}{3}\)

 

Because two tangent lines of a circle intersect at an external point, J, \(\angle IJB\cong\angle HJB\). Using the same reasoning as before, \(\tan m\angle IJB=\tan m\angle HJB=\frac{4}{3}\)\(m\angle IJH=m\angle IJB+m\angle HJB=m\angle HJB+m\angle HJB=2m\angle HJB\)

 

Both tangent lines meet at an angle of \(2m\angle HJB\). Just like finding \(\tan m\angle KAB\) yielded the slope, \(\tan 2m\angle HJB\) will yield the slope of the desired line, \(\overleftrightarrow{IJ}\)

 

Luckily, we can utilize the Double-Angles Identities, specifically the \(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB}\).

\(\tan 2m\angle HJB=\frac{2\tan m\angle HJB}{1-\tan^2 m\angle HJB};\\ \tan m\angle HJB=\frac{4}{3}\) This is the Double-Angle Identity for the tangent function. Replace all instances of \(\tan m\angle HJB\) with 4/3.
\(\tan 2m\angle HJB=\frac{2*\frac{4}{3}}{1-\left(\frac{4}{3}\right)^2}\\ \tan 2m\angle HJB=\frac{\frac{8}{3}}{\frac{-7}{9}}\\ \tan 2m\angle HJB=\frac{-24}{7} \) Simplify. The result is the slope of the line \(\overleftrightarrow{IJ}\) .
\(y=\frac{-24}{7}x+b\) This is the equation of the line \(\overleftrightarrow{IJ}\).

 

The equation of the circle with the center A is \((x-2)^2+(y-1)^2=1\). Since \(y=\frac{-24}{7}x+b\), substitute that into the equation for the circle:

 

\((x-2)^2+(y-1)^2=1;\\ y=\frac{-24}{7}x+b \)

These are the equations we are grappling with.
\((x-2)^2+(\frac{-24}{7}x+[b-1])^2=1\) It is time to expand this monstrosity.
\(x^2-4x+4+\left(\frac{-24}{7}x\right)^2+2*\frac{-24}{7}x*(b-1)+(b-1)^2=1\)  
\(x^2-\frac{28}{7}x+4+\frac{576}{49}x^2-\frac{48b-48}{7}x+b^2-2b+4=0\)  
\(\textcolor{red}{\frac{625}{49}}x^2\textcolor{blue}{-\left(\frac{48b-20}{7}\right)}x+\textcolor{green}{(b^2-2b+4)}=0\\ \textcolor{red}{a}x^2+\textcolor{blue}{b}x+\textcolor{green}{c}=0\) Notice the parallelism between a quadratic and the current equation. Tangent lines always intersect a circle at only one point, so let's set the discriminant equal to zero to ensure only one solution is possible.
\(\Delta=b^2-4ac;\\ \Delta=\left(-\frac{48b-20}{7}\right)^2-4*\frac{625}{49}*(b^2-2b+4)\\ \Delta=\frac{-196b^2+3080b-9600}{49}\) Set the discriminant equal to zero and solve.
\(\frac{-196b^2+3080b-9600}{49}=0\Rightarrow-196b^2+3080b-9600=0\) Use the quadratic formula here.
\(b_1=\frac{30}{7}\text{ or }b_2=\frac{80}{7}\) Reject b_2 because that is on the other side of the circle.