The locus of the mid-points of chords of the circle x2 + y2 = 4 which subtends a right angle at the origin is

Question

The locus of the mid-points of chords of the circle x2 + y2 = 4 which subtends a right angle at the origin is (A) x + y = 2  (B) x2 + y2 = 1  (C)  x2 + y2 = 2  (D) x + y = 1.

Tardigrade Admin · Accepted Answer

We have to find locus of mid-point of chord and we know perpendicular from centre bisects the chord.

∴ ∠ O A C = 4 5^{\circ}

In

△ O A C, \frac{OC}{O A} = sin 4 5^{\circ} \Rightarrow OC = \frac{2}{2} = 2

Also,

h^{2} + k^{2} = O C^{2}

Hence,

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

is required equation of locus of mid-point of chord subtending right angle at the centre.

Q. The locus of the mid-points of chords of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

$x + y = 2$

$x^{2} + y^{2} = 1$

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

$x + y = 1.$

Q. The locus of the mid-points of chords of the circle x2+y2=4 which subtends a right angle at the origin is

x+y=2

x2+y2=1

x2+y2=2

x+y=1.

We have to find locus of mid-point of chord and we know perpendicular from centre bisects the chord. ∴∠OAC=45∘ In △OAC,OAOC​=sin45∘⇒OC=2​2​=2​ Also, h2+k2​=OC2 Hence, x2+y2=2 is required equation of locus of mid-point of chord subtending right angle at the centre.

Q. The locus of the mid-points of chords of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

$x + y = 2$

$x^{2} + y^{2} = 1$

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

$x + y = 1.$