coordinate geometry

Feb 2010
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find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0
 

earboth

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find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0
1. The circle \(\displaystyle c_1: x^2+y^2=4^2\) has the center (0, 0) and the radius 4.
The circle \(\displaystyle c_2:x^2+y^2-12x+32=0~\implies~(x-6)^2+y^2=2^2\) has the center (6, 0) and the radius 2.

2. Use similar right triangles to determine the missing lengthes. I used Euklid's theorem to calculate the coordinates of the tangent points: \(\displaystyle T_1\left(\frac43\ ,\ \frac83 \sqrt{2}\right)\)

and \(\displaystyle T_2\left(\frac{20}3\ ,\ \frac43 \sqrt{2}\right)\)

3. The center of the resulting circle \(\displaystyle c_r\) must be on the x-axis and on the perpendicular bisector of \(\displaystyle \overline{T_1T_2}\)
The slope of this perpendicular bisector equals the slope of \(\displaystyle \overline{OT_1} = \dfrac{\frac83 \sqrt{2}}{\frac43}=2\sqrt{2}\)

The midpoint of \(\displaystyle \overline{T_1T_2}\) has the coordinates \(\displaystyle M\left(\dfrac{\frac43 + \frac{20}3}2\ , \ \dfrac{\frac83 \sqrt{2} + \frac{4}3 \sqrt{2}}2 \right)\). That means \(\displaystyle M(4, 2\sqrt{2})\)

4. The perpendicular bisector through M has the equation:

\(\displaystyle y-2 \sqrt{2} = 2\sqrt{2}(x-4)\)

This line crosses the x-axis at:

\(\displaystyle -2 \sqrt{2} = 2\sqrt{2}(x-4)~\implies~-1=x-4~\implies~x=3\)

5. The center of the resulting circle is C(3, 0). Now calculate the distance \(\displaystyle |\overline{CT_1}|\) to get the radius of \(\displaystyle c_r\). I've got \(\displaystyle r=\sqrt{17}\). Therefore the equation of \(\displaystyle c_r\) is:

\(\displaystyle c_r: (x-3)^2+y^2=17\)
 

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Nov 2007
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Thanks earboth a nice solution

bjh
 
Feb 2010
460
37
how did yu use euklids geometry here and what is it basically
 

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
how did yu use euklids geometry here and what is it basically
1. I learned that in English Euclid's theorem is related to number theory. In Germany a group of theorems is summed under the name Euclid's theorems where this theorem belongs to:

In a right triangle the height divides the hypotenuse into two segments, p and q. Then

\(\displaystyle h^2 = p \cdot q\)

And that is what I used:

2. Take \(\displaystyle c_1\) and the corresponding right triangle with

\(\displaystyle c = 12, b = 4, a = 8 \sqrt{8}\)

According to the the attached sketch the tangent point \(\displaystyle T_1\) has the coordinates \(\displaystyle T_1(x, h)\)

3. From the area of the right tringle you know:

\(\displaystyle A=\frac12 \cdot c \cdot h = \frac12 \cdot a \cdot b ~\implies~h = \dfrac{a \cdot b}{c}\)

4. With y = 12 - x you get:

\(\displaystyle x(12-x)=h^2\)

Since you know already the value of h you can solve the equation for x.

5. Alternatively you can plug in the value of h² instead of y² into the equation of the circle to get the x-coordinate of the tangent point:

\(\displaystyle x^2+\dfrac{16 \cdot 128}{144} = 16~\implies~ |x|=\dfrac34\)
 

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