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

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- Jul 18th 2010, 07:07 AMprasumcoordinate 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, 12:02 PMearboth
1. The circle has the center (0, 0) and the radius 4.

The circle 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:

and

3. The center of the resulting circle must be on the x-axis and on the perpendicular bisector of

The slope of this perpendicular bisector equals the slope of

The midpoint of has the coordinates . That means

4. The perpendicular bisector through M has the equation:

This line crosses the x-axis at:

5. The center of the resulting circle is C(3, 0). Now calculate the distance to get the radius of . I've got . Therefore the equation of is:

- Jul 18th 2010, 12:16 PMbjhopper
Thanks earboth a nice solution

bjh - Jul 19th 2010, 03:18 AMprasum
thanks man

- Jul 20th 2010, 06:50 PMprasum
how did yu use euklids geometry here and what is it basically

- Jul 20th 2010, 11:34 PMearboth
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

And that is what I used:

2. Take and the corresponding right triangle with

According to the the attached sketch the tangent point has the coordinates

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

4. With y = 12 - x you get:

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: