# Question #6de4a

Jul 23, 2016

The locus of all such points is a circle of radius $\sqrt{2}$ with a center at origin. The equation describing this circle is
(3) ${x}^{2} + {y}^{2} = 2$

#### Explanation:

The circle with equation ${x}^{2} + {y}^{2} = 4$ has a center at origin and radius $R = 2$.

All chords sub-tending a right angle in this circle (as all chords sub-tending a fixed angle in any circle) have fixed length.

A triangle formed by a chord sub-tending the right angle and two radii to its ends is a right isosceles triangle with a chord as a hypotenuse and each radius as a cathetus.

Therefore, the length of any chord, sub-tending the right angle in a circle of a radius $R = 2$, by Pythagorean Theorem has the length
$L = \sqrt{{R}^{2} + {R}^{2}} = 2 \sqrt{2}$

The segment from the center of a circle to a midpoint of a triangle described above is an altitude, median and right angle bisector of this right isosceles triangle. The length of this segment can also be calculated using Pythagorean Theorem as
$H = \sqrt{{R}^{2} - {\left(\frac{L}{2}\right)}^{2}} = \sqrt{4 - {\left(\sqrt{2}\right)}^{2}} = \sqrt{2}$

This length $H$ is a distance from the origin to a midpoint of ANY chord sub-tending a right angle in our circle.

Therefore, the locus of all such points is a circle of radius $H = \sqrt{2}$ with a center at origin. The equation describing this circle is
(3) ${x}^{2} + {y}^{2} = 2$

Nov 13, 2017

The Right Choice is Option $\left(3\right) : {x}^{2} + {y}^{2} = 2.$

#### Explanation:

Observe that, ${x}^{2} + {y}^{2} = 4$ represents a circle, having centre at the

Origin $O = O \left(0 , 0\right)$ and radius $r = 2.$

The parametric eqns. of this circle are,

$x = 2 \cos \theta , y = 2 \sin \theta , \theta \in \left[0 , 2 \pi\right) .$

Let $P = P \left(2 \cos \theta , 2 \sin \theta\right)$ be one end-point of a chord $P Q$

of this circle.

Then, $\vec{O P}$ makes an $\angle \theta$ with the $+ v e$ direction of

the X-axis.

Since, $P Q$ subtends a right angle at the Origin $O ,$ we find

that $\vec{O Q}$ must be making an $\angle \left(\theta \pm \frac{\pi}{2}\right)$ with the

$+ v e$ direction of the X-axis.

$\therefore Q = Q \left(2 \cos \left(\theta \pm \frac{\pi}{2}\right) , 2 \sin \left(\theta \pm \frac{\pi}{2}\right)\right) , i . e . ,$

$Q = Q \left(- 2 \sin \theta , 2 \cos \theta\right) , \mathmr{and} , Q = Q \left(2 \sin \theta , - 2 \cos \theta\right) .$

If $Q = Q \left(- 2 \sin \theta , 2 \cos \theta\right) , \mathmr{and} , P = P \left(2 \cos \theta , 2 \sin \theta\right) ,$

then, the mid-point $M \text{ of } P Q$ is $M \left(\cos \theta - \sin \theta , \cos \theta + \sin \theta\right) .$

To find the Locus of M, let us set, $M = M \left(X , Y\right)$, then, we have,

$X = \cos \theta - \sin \theta , Y = \cos \theta + \sin \theta .$

Eliminating $\theta$ from this, we get, ${X}^{2} + {Y}^{2} = 2 ,$ showing that

the general eqn. of the desired locus is ${x}^{2} + {y}^{2} = 2.$

In case, $Q = Q \left(2 \sin \theta , - 2 \cos \theta\right) ,$ we get the same result.

Hence, the right choice is option $\left(3\right) : {x}^{2} + {y}^{2} = 2.$

Enjoy Maths.!