anonymous
  • anonymous
Fool's problems of the day, Today's problems are based on standard analytic geometry, \((1)\) Find the center of the conic section \(13x^2-18xy+37y^2+2x+14y-2=0 \) [ Solved by @Mani_Jha and @.Sam. ] \((2)\) Prove that if the circles \(x^2+y^2+2g_1x+2f_1y+c_1 =0 \) and \(x^2+y^2+2g_2x+2f_2y+c_2 =0 \) are orthogonal then \( 2(g_1g_2 +f_1f_2) = c_1+c_2 \) [Solved by @Ishaan94]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Mani_Jha
  • Mani_Jha
The first one can be found by : 1) Partially differentiating the function with respect to x, and then setting the result to zero. This will give the x-coordinate. 2)Partially differentiating the function with respect to y, and then setting the result to zero. This will give the y-coordinate. Is that right?
anonymous
  • anonymous
@Mani_Jha: Right, do you know why this works? ;)
Mani_Jha
  • Mani_Jha
Atcually, no :\.

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anonymous
  • anonymous
Let me guess, ripped from IIT/AIEEE prep module? ;)
Mani_Jha
  • Mani_Jha
I read it somewhere. But I didn't know the reason. Can you tell me? The second one can be solved by writing the tangent equation for one circle, normal equation for the other and equating them. (I guess that's the meaning of orthogonal)
anonymous
  • anonymous
I'm interested to know why, too :P
anonymous
  • anonymous
I think this explains why... |dw:1333379044194:dw|
anonymous
  • anonymous
nice
.Sam.
  • .Sam.
I have the solution for (1)
.Sam.
  • .Sam.
but without working
anonymous
  • anonymous
without working?
anonymous
  • anonymous
@ishaan94
anonymous
  • anonymous
What do you mean by orthogonal? I always forget its meaning :/
anonymous
  • anonymous
|dw:1333381204296:dw|
.Sam.
  • .Sam.
i know the solution for (1) let me type it out
anonymous
  • anonymous
And if anybody interested, this is the problem that's bugging me: http://math.stackexchange.com/questions/127252/ :((((((
anonymous
  • anonymous
2 is really easy
anonymous
  • anonymous
Sure Ishaan I always post easy problems :)
anonymous
  • anonymous
Ishaan post explanation please :)
.Sam.
  • .Sam.
13x^2−18xy+37y^2+2x+14y−2=0 \[\frac{\partial d}{\partial x}=26x-18y+2=0\] \[\frac{\partial d}{\partial y}=-18x+74y+2=0\] -18x+74y+2=0...............(1) 26x-18y+2=0................(2) Simultaneous and get (-0.25,-0.25)
anonymous
  • anonymous
Firstset, \(d= 13x^2−18xy+37y^2+2x+14y−2=0\)
anonymous
  • anonymous
|dw:1333381399174:dw|Two circles with centers \((-g_1,-f_1)\) and \((-g_2,-f_2)\). \[r_1 = \sqrt{g_1^2 +f_1^2 - c_1}\]\[r_2 = \sqrt{g_2^2 +f_2^2 - c_2}\] Apply Pythagoras theorem, \[(g_1 -g_2)^2 + (f_1-f_2)^2 = g_1^2 +f_1^2 - c_1+g_2^2 +f_2^2 - c_2\]
anonymous
  • anonymous
No, not always maybe not always easy for me...
anonymous
  • anonymous
Can someone explain why partial derivatives works? I only know the method, I have no idea why the method works and how it works.
anonymous
  • anonymous
Amazing ishaan :D I am assuming you know the general expression for any angle \( \theta \) right?
anonymous
  • anonymous
I still don't understand how the distance between the circles relates to orthogonality :(
anonymous
  • anonymous
Tangents are perpendicular, and a line perpendicular to tangent passing through the point of contact satisfies the circle's center. That's all I used. @AsianDuck I think I do FoolForMath
anonymous
  • anonymous
Thank you :)
anonymous
  • anonymous
"Tangents are perpendicular" Tangents to the two circle at the point of their intersection are perpendicular.
anonymous
  • anonymous
Anyways, it's given by \[ \cos \theta = \frac{r_1^2+r_2^2 -d^2} {2r_1r_2}\] Where d is the distance between the centers. Others can try proving it too ;) PS: Super easy !

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