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  
Global Moderator Joined: May 2007 Posts: 6,787 Thanks: 708  Re: Circle Quote:
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 (xxc)^2 + (yyc)^2 = r^2, where r is the radius and (xc,yc) is the center.  
October 20th, 2012, 06:13 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  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 xcoordinates of the endpoints of the diameter of the circle. Substitute into [color=blue][1][/color] and determine the corresponding ycoordintes. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ [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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