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Prove that the locus of the midpoints of the chord of the circle $x^2+y^2-2x+2y-2=0$ parallel to the line $y=x+5$ is the line which passes through (0,0)

Let the point be (h,k)

$$T=S_1$$ $$xx_1+yy_1+g(x+x_1)+f(y+y_1)-c=x_1^2+y_1^2+2gx_1+2fy_1-c$$ where $(x_1,y_1)$ are the given midpoints of the chord. $$hx+ky+(-1)(x+h)+(1)(y+k)-2=h^2+k^2-2h+2k-2$$

We only have to deal with terms containing x and y $$x(h-1)+y(k+1)+\lambda=0$$ Since it is parallel to $x-y+5=0$ $$h-1=1$$ and $$k+1=-1$$

$h=2$ and $k=-2$

What is going wrong?

Aditya
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    Note that no calculation is needed. The centre lies on the line $x+y=0$ and the chords are all perpendicular to that line, so it passes through their midpoints. – almagest Jan 26 '20 at 09:57
  • @almagest I don’t follow. How can the center lie on a single line, since it passes through infinite lines? How did you arrive at $x+y=0$? And what’s wrong with what I did? – Aditya Jan 26 '20 at 10:42
  • It is a single circle, so it has a single centre! – almagest Jan 26 '20 at 10:44
  • @almagest a single center from which infinite lines can pass – Aditya Jan 26 '20 at 10:50
  • but only one perpendicular to $y=x+5$. – almagest Jan 26 '20 at 10:53
  • Incidentally, I have no idea what you did. Presumably you are taking $(h,k)$ to be the midpoint of a chord, but what is the lhs of the equation following $T=S_1$ supposed to be? And what are $T,S_1$? – almagest Jan 26 '20 at 10:56
  • Okay, forgive me for thinking it was standard. I will add it – Aditya Jan 26 '20 at 10:58

1 Answers1

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It should be slope of two lines be same. $$m=-\frac{h-1}{k+1}=1$$ Hence $-h+1=k+1$ Hence locus is $ x+y=0$

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