Find the radius of the smaller circle in the tangent problem

In summary, the conversation discusses the solution to a problem involving three circles and their radii. Two approaches are presented, one using Pythagoras' theorem and another using Cartesian coordinates. The summary concludes that the solution to the problem is ##c=6-4\sqrt{2}##.
  • #1
chwala
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Homework Statement
see attached
Relevant Equations
pythagoras theorem
Find the question here and the solution i.e number 10 indicated as ##6-4\sqrt{2}##,

1647588780045.png


1647588821782.png

I am getting a different solution, my approach is as follows. I made use of pythagoras theorem for the three right angle triangles as follows,
Let radius of the smaller circle be equal to ##c## and distance from mid point of smaller circle to circle with radius ##1=x## then it follows that,
##x^2+(1-c)^2=(1+c)^2##
##x^2=4c##

let us also have the distance from the smaller circle to circle with radius ##2=y##, then it follows that,
##y^2+(2-c)^2=(2+c)^2##
##y^2=8c##

Let ##y^2+x^2=AB##, then it follows that, ##(AB)^2 + (2-1)^2=3^2##
##AB=2\sqrt{2}##
then it follows that, ##\sqrt{8}=2\sqrt{c}+2\sqrt{2c}##

Am i missing something here!

i got it!
##\sqrt{c}##=##\dfrac {\sqrt {8}}{2+2\sqrt{2}}##
⇒##c=0.34314575## Bingo guys! Phew i took time on this men!:cool:
 
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  • #2
Another approach by introducing Cartesian coordinates.
Equation of circle blue
[tex]x^2+y^2=2^2[/tex]
Equation of circle green
[tex](x-3)^2+y^2=1^2[/tex]
Equation of common tangent line
[tex]y=\frac{1}{2\sqrt{2}}(x-6)[/tex]
Let equation of circle red be
[tex]x^2+y^2+2px+2qy=r [/tex]
which is tangential to all these three figures. Thus we can have three linear equations of p, q and r, and get
[tex]c=\sqrt{r+p^2+q^2}[/tex]

More simply from the figure
[tex]2\tan^{-1}\frac{c}{4}=tan^{-1}\frac{1}{2\sqrt{2}}[/tex]
[tex]c= 4 \tan (\frac{1}{2}tan^{-1}\frac{1}{2\sqrt{2}})=4\tan (\frac{sin \theta}{1+cos \theta})[/tex]
where ##\theta=tan^{-1}\frac{1}{2\sqrt{2}}##
[tex]c=4\tan \frac{1/3}{1+2\sqrt{2}/3}=4\tan (3-2\sqrt{2})\approx 0.012[/tex]
I should appreciate if you would check whether it works or not.
 
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  • #3
anuttarasammyak said:
Another approach by introducing Cartesian coordinates.
Equation of circle blue
[tex]x^2+y^2=2^2[/tex]
Equation of circle green
[tex](x-3)^2+y^2=1^2[/tex]
Equation of common tangent line
[tex]y=\frac{1}{2\sqrt{2}}(x-6)[/tex]
Let equation of circle red be
[tex]x^2+y^2+2px+2qy=r [/tex]
which is tangential to all these three figures. Thus we can have three linear equations of p, q and r, and get
[tex]c=\sqrt{r+p^2+q^2}[/tex]

More simply from the figure
[tex]2\tan^{-1}\frac{c}{4}=tan^{-1}\frac{1}{2\sqrt{2}}[/tex]
[tex]c= 4 \tan (\frac{1}{2}tan^{-1}\frac{1}{2\sqrt{2}})=4\tan (\frac{sin \theta}{1+cos \theta})[/tex]
where ##\theta=tan^{-1}\frac{1}{2\sqrt{2}}##
[tex]c=4\tan \frac{1/3}{1+2\sqrt{2}/3}=4\tan (3-2\sqrt{2})\approx 0.012[/tex]
I should appreciate if you would check whether it works or not.
[tex]c=4\tan \frac{1/3}{1+2\sqrt{2}/3}=4\tan (3-2\sqrt{2})\approx 0.012[/tex]I am afraid this does not look correct, i overlooked your solution. I have just checked it in detail...
 
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  • #4
I made a mistake. Thanks for your effort.
 
  • #5
I was thinking of using linear scale factor...i do not know whether that's possible. i'll try check it out...
 
  • #6
Let me try again. Observing angles which the triangle whose vertexes are center of circles and horizontal line on which center of orange circle lies make,
[tex]\alpha+\beta+\gamma=\pi[/tex]
where
[tex]\sin\alpha=\frac{2-c}{2+c}[/tex]
[tex]\sin\gamma=\frac{1-c}{1+c}[/tex]
[tex]3^2=(2+c)^2+(1+c)^2-2 (2+c)(1+c)\cos\beta[/tex]
This leads to ##c=6-4\sqrt{2}## which is same as OP.
 
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FAQ: Find the radius of the smaller circle in the tangent problem

1. How do I find the radius of the smaller circle in the tangent problem?

To find the radius of the smaller circle in the tangent problem, you will need to use the Pythagorean theorem. First, draw a line from the center of the larger circle to the point where the smaller circle is tangent to it. This line will be the hypotenuse of a right triangle. Then, use the radius of the larger circle as one leg of the triangle and the distance between the centers of the two circles as the other leg. Finally, use the Pythagorean theorem (a^2 + b^2 = c^2) to solve for the radius of the smaller circle.

2. What is the Pythagorean theorem?

The Pythagorean theorem is a mathematical formula that relates the three sides of a right triangle. It states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

3. Can I use any other method to find the radius of the smaller circle?

Yes, there are other methods that can be used to find the radius of the smaller circle in the tangent problem. One method is to use the formula for the distance between two points in a coordinate plane. Another method is to use trigonometric functions such as sine, cosine, and tangent. However, the Pythagorean theorem is often the simplest and most straightforward method to use.

4. What information do I need to know in order to find the radius of the smaller circle?

In order to find the radius of the smaller circle, you will need to know the radius of the larger circle and the distance between the centers of the two circles. These two pieces of information are crucial for using the Pythagorean theorem to solve for the radius of the smaller circle.

5. Can I use this method for any tangent problem involving circles?

Yes, the method of using the Pythagorean theorem to find the radius of the smaller circle can be applied to any tangent problem involving circles. As long as you have the necessary information (radius of the larger circle and distance between the centers of the two circles), you can use this method to solve for the radius of the smaller circle.

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