Question

Locus of the mid-points of the chords of the circle $x^2+y^2=4$ which subtend a right angle at the centre is

• A. $x+y=2$
• B. $x^2+y^2=1$
• C. $x^2+y^2=2$
• D. $x-y=0$

Answer 1

The correct option is C $x^2+y^2=2$

Let AB = a

$⟹AC=CB=\frac{a}{2}$

Given that $\angle AOB={90}^{\circ }$

$⟹A{B}^2=O{A}^2+O{B}^2$

$a^2=r^2+r^2=2r^2=2\times 4=8$

AB is any chord and $C(x_1,y_1)$ be the midpoint of AB

In $△OAC$

$OC\perp AC$

$⟹O{C}^2=O{A}^2–A{C}^2$

$x_1^2+y_1^2=r^2-\frac{a^2}{4}$

$x_1^2+y_1^2=r^2-\frac{2r^2}{4}=\frac{r^2}{2}=\frac{4}{2}=2$