Question

Question #6de4a

Answer 1

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) = 2sqrt(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-(L/2)^2) = sqrt(4-(sqrt(2))^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`

Answer 2

The Right Choice is Option `(3) : x^2+y^2=2.`

Explanation:

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

Origin `O=O(0,0)` and radius `r=2.`

The parametric eqns. of this circle are,

`x=2costheta, y=2sintheta, theta in [0,2pi).`

Let `P=P(2costheta,2sintheta)` be one end-point of a chord `PQ`

of this circle.

Then, `vec(OP)` makes an `/_theta` with the `+ve` direction of

the X-axis.

Since, `PQ` subtends a right angle at the Origin `O,` we find

that `vec(OQ)` must be making an `/_(theta+-pi/2)` with the

`+ve` direction of the X-axis.

`:. Q=Q(2cos(theta+-pi/2),2sin(theta+-pi/2)), i.e.,`

`Q=Q(-2sintheta,2costheta), or, Q=Q(2sintheta,-2costheta).`

If `Q=Q(-2sintheta,2costheta), and, P=P(2costheta,2sintheta),`

then, the mid-point `M" of "PQ` is `M(costheta-sintheta,costheta+sintheta).`

To find the Locus of M, let us set, `M=M(X,Y)`, then, we have,

`X=costheta-sintheta, Y=costheta+sintheta.`

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(2sintheta,-2costheta),` we get the same result.

Hence, the right choice is option `(3) : x^2+y^2=2.`

Enjoy Maths.!