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  • The locus of the mid point of a chord of the circle , which subtends a

The locus of the mid point of a chord of the circle \mathrm{x^2+v^2=4}, which subtends a right angle at the origin is

Option: 1

x+y=2


Option: 2

x^2+y^2=1


Option: 3

x^2+y^2=2


Option: 4

x+y=1


Answers (1)

best_answer

As, we have to find locus of mid-point of chord and

we know perpendicular from centre bisects the chord.

So, chord joining A(r, 0) and B(0, r) subtends a right angle

at the centre (0,0). Mid point of AB is \left(\frac{r}{2}, \frac{r}{2}\right).

\therefore O C=\frac{r}{\sqrt{2}}, which is radius of locus of C.

Posted by

Suraj Bhandari

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