find locus of midpoint of all the chords of circle $$x^2+ y^2- 2x- 2y=0$$ such that pair of line joining (0,0) and the point of intersection of the chords with circles make equal angles with X axis.
First of all I took the midpoint of chord as (h, k) and one point on a variable chord( satisfying the given condition) as well as the circle as ($\alpha,\beta$). So the other point on chord as well as the circle will be ($2h-\alpha, 2k-\beta$). Using some coordinate geometry I got three equations $$(\alpha) ^2-(\beta)^2=2h\alpha -2k\beta$$ $$(\alpha) ^2+(\beta)^2=2(\alpha+\beta) $$ $$h^2+k^2+h+k=h\alpha+k\beta$$
But now could not proceed further. Thanks in advance.
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asked Oct 10, 2017 at 6:36
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1 Answer
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Let the midpoint of chord be $(h,k)$
Then equation of chord is $$xh+yk-(x+h)-(y+k)=h^2+k^2-2h-2k$$ now equation of pair of straight line joining origin to point of contact of the line with the circle is made by homogenising the equation of circle.Thus the equation of pair of straight lines become $$x^2+y^2-(x+y)\frac{(xh+yk-x-y)}{h^2+k^2-h-k}=0$$ Now since this makes equal angle with x axis implies $m_1+m_2=0$ or $-2\frac{(-k+1-h+1)}{1-k}=0$ So $x+y=2$ is the desired locus.
