# coordinate geometry

• Jul 18th 2010, 06:07 AM
prasum
coordinate geometry
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
• Jul 18th 2010, 11:02 AM
earboth
Quote:

Originally Posted by prasum
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:

$\displaystyle c_r: (x-3)^2+y^2=17$
• Jul 18th 2010, 11:16 AM
bjhopper
Thanks earboth a nice solution

bjh
• Jul 19th 2010, 02:18 AM
prasum
thanks man
• Jul 20th 2010, 05:50 PM
prasum
how did yu use euklids geometry here and what is it basically
• Jul 20th 2010, 10:34 PM
earboth
Quote:

Originally Posted by prasum
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:

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