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October 20th, 2012, 08:40 AM   #1
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Circle

Hey guys please help in understanding this
find the equation of circle joining points of intersection of ax^2 +by^2 +2hxy=0
lx+my =1 as diameter
Raghav C Marthi is offline  
 
October 20th, 2012, 01:01 PM   #2
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Re: Circle

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Originally Posted by Raghav C Marthi
Hey guys please help in understanding this
find the equation of circle joining points of intersection of ax^2 +by^2 +2hxy=0
lx+my =1 as diameter
Step 1: Find points of intersect use lx + my = 1 to get y= or x = and substitute into first equation to get a quadratic (be careful if m = 0 or l = 0, both can't be).

Step 2: Solve quadatic and substitute into lx+my=1 to get other ordinate. You now have the end points (x1,y1) and (x2,y2).

Step 3: The midpoint is the center of the circle and the radius is half the distance between the end points.

Your circle is (x-xc)^2 + (y-yc)^2 = r^2, where r is the radius and (xc,yc) is the center.
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October 20th, 2012, 06:13 PM   #3
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Re: Circle

Hello, Raghav C Marthi!

Who assigned this problem?
It is one of the most sadistic that I've ever seen!


Quote:

I get a few steps into the solution . . . and I must quit!

[color=beige] .[/color][color=blue][1][/color]




[color=beige] .[/color][color=red]**[/color]




[color=beige] .[/color][color=red]**[/color]


We have the x-coordinates of the endpoints of the diameter of the circle.
Substitute into [color=blue][1][/color] and determine the corresponding y-coordintes.







~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[color=red]**[/color]
I don't guarentee the algebra in these two steps.

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October 21st, 2012, 08:00 AM   #4
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Re: Circle

Hey dude Thank you
Raghav C Marthi is offline  
October 21st, 2012, 08:02 AM   #5
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Re: Circle

Thank you mathman and soroban... helped me a lot
Raghav C Marthi is offline  
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