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?
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$