Question

The locus of the mid-point of a chord of the circle $x^2+ y^2 = 4$, which subtends a right angle at the origin is

• A. $x + y = 2$
• B. $x^{2}+y^{2}=1$
• C. $x^{2}+y^{2}=2$
• D. $x + y = 1$

Answer 1

Answer: (C)

$x^{2}+y^{2}=2$

Answer 2

we have given circle $x^{2}+y^{2}=4$. so, $r=2$. So, radius will be $2$.

we have given circle
we know that, Perpendicular from center i.e. $OM$ bisects the chord $A B$.
$\therefore \angle O A M=45{\circ}$
$\therefore \sin 45^{\circ}=\frac{O M}{A M}$
$\Rightarrow O M=\sin 45^{\circ} \cdot A M$
$\Rightarrow O M=\frac{1}{\sqrt{2}} \times 2$ (AM is radius)
$\Rightarrow O M=\sqrt{2}$
$\Rightarrow O M^{2}=2$
$\Rightarrow h^{2}+k^{2}=O M^{2}$
$\Rightarrow h^{2}+k^{2}=2$
Hence, the focus of $(h, k)$ is $x^{2}+y^{2}=2$.