1. ## 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

2. 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 $c_1: x^2+y^2=4^2$ has the center (0, 0) and the radius 4.
The circle $c_2:x^2+y^2-12x+32=0~\implies~(x-6)^2+y^2=2^2$ 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: $T_1\left(\frac43\ ,\ \frac83 \sqrt{2}\right)$

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

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

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

4. The perpendicular bisector through M has the equation:

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

This line crosses the x-axis at:

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

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

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

3. Thanks earboth a nice solution

bjh

4. thanks man

5. how did yu use euklids geometry here and what is it basically

6. 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

$h^2 = p \cdot q$

And that is what I used:

2. Take $c_1$ and the corresponding right triangle with

$c = 12, b = 4, a = 8 \sqrt{8}$

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

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

$A=\frac12 \cdot c \cdot h = \frac12 \cdot a \cdot b ~\implies~h = \dfrac{a \cdot b}{c}$

4. With y = 12 - x you get:

$x(12-x)=h^2$

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:

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