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 October 20th, 2012, 08:40 AM #1 Newbie   Joined: Oct 2012 Posts: 7 Thanks: 0 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
October 20th, 2012, 01:01 PM   #2
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Re: Circle

Quote:
 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.

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:
 $\text{Find the equation of circle joining points of intersection of} \;\;\;ax^2\,+\,by^2\,+\,2hxy\:=\:0 \text{ and } lx\,+\,my \:=\:1\text{ as diameter.}$

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

$\text{Solve the linear equation for }y:\;\;y \:=\:\frac{1\,-\,lx}{m}$[color=beige] .[/color][color=blue][1][/color]

$\text{Substitute into the quadratic: }\:ax^2\,+\,b\left(\frac{1\,-\,lx}{m}\right)^2\,+\,2hx\left(\frac{1\,-\,lx}{m}\right) \;=\;0$

$\;\;\;\text{which simplifies to: }\:(am^2\,+\,bl^2\,-\,2hlm)x^2\,+\,2(hm\,-\,bl)x\.+\,b \;=\;0$[color=beige] .[/color][color=red]**[/color]

$\text{Quadratic Formula: }\:x \;=\;\frac{2(bl\,-\,hm)\,\pm\,\sqrt{4(hm-bl)^2\,-\,4b(am^2\,+\,bl^2\,-\,2hlm)}}{2(am^2\,+\,bl^2\,-\,2hlm)}$

$\;\;\;\text{which simplifies to: }\:x \;=\;\frac{(bl\,-\,hm)\,\pm\,m\sqrt{h^2\,-\,ab}}{am^2\,+\,bl^2\,-\,2hm}$[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.

$\text{Then we have the two points: }\,P(x_1,\,y_1)\,\text{ and }\,Q(x_2,\,y_2)$

$\text{The center of the circle is the midpoint of }P\text{ and }Q:\;C\,\left(\frac{x_1\,+\,x_2}{2},\:\frac{y_1\,+ \,y_2}{2}\right)$
$\;\;\;\text{and the radius is half the distance }\overline{PQ}.$

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

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

 October 21st, 2012, 08:00 AM #4 Newbie   Joined: Oct 2012 Posts: 7 Thanks: 0 Re: Circle Hey dude Thank you
 October 21st, 2012, 08:02 AM #5 Newbie   Joined: Oct 2012 Posts: 7 Thanks: 0 Re: Circle Thank you mathman and soroban... helped me a lot

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