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```
the equation of the locus of the middle point of a chord of the circle x^2 + y^2=2(x+y) such that the pair of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the x-axis is:

(a)x+y=2
(b)x-y=2
(c)2x-y=1
(d)none of these

I WILL B VERY THANKFUL TO U IF U GIVE AN ELABORATED SOLUTION...THANKU
```
7 years ago

510 Points
```										let one of the line is y=mx                   (line is passing through origin)
this line makes an angle @ with x axis ...another line is also inclined to x axis with same angle but its slope will be -m..
now other will be y=-mx
let point of contact of line y=mx & y=-mx with circle be P & q respectively then
on solving y=mx and circle we get
P=[  2(1+m)/1+m2   ,   2m(1+m)/1+m2  ]
and on solving y=-mx and circle we get
q=[  2(1-m)/1+m2   ,   2m(m-1)/1+m2   ]
we have to find the locus of mid point of PQ because pq is chord of circle,
let X,Y is the mid point then
X=2/1+m2  &  Y=2m2 /1+m2
on solving these we get
X+Y=2
THIS IS THE REQUIRED EQUATION
```
7 years ago
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