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]

Question

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]

mani_jha · Answer

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 · Answer

@Mani_Jha: Right, do you know why this works? ;)

mani_jha · Answer

Atcually, no :\.

anonymous · Answer

Let me guess, ripped from  IIT/AIEEE prep module? ;)

mani_jha · Answer

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 · Answer

I'm interested to know why, too :P

anonymous · Answer

I think this explains why...

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