The locus of the midpoint of a chord of the c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the midpoint be $(h, k)$. The equation of the chord with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$.Homogenizing the circle equation $x^2 + y^

Vedclass · Accepted Answer

$x^2 + y^2 = 2$

Vedclass · Answer

(C) Let the midpoint be $(h, k)$. The equation of the chord with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$.Homogenizing the circle equation $x^2 + y^2 = 4$ using the chord equation:$x^2 + y^2 = 4 \left( \frac{hx + ky}{h^2 + k^2} \right)^2$.Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero.$1 - \frac{4h^2}{(h^2 + k^2)^2} + 1 - \frac{4k^2}{(h^2 + k^2)^2} = 0$.$2 - \frac{4(h^2 + k^2)}{(h^2 + k^2)^2} = 0$.$2 - \frac{4}{h^2 + k^2} = 0$.$h^2 + k^2 = 2$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = 2$.

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

Similar Questions

Let $C$ be the circle with centre $(0,0)$ and radius $3$ units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2\pi}{3}$ at its centre is:

The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

If the angle between a pair of tangents drawn from a point $P$ to the circle $x^2+y^2+4x-6y+9 \sin^2 \alpha+13 \cos^2 \alpha=0$ is $2 \alpha$,then the equation of the locus of $P$ is

If a bar of given length moves with its extremities on two fixed straight lines at right angles,then the locus of any point marked on the bar describes a/an

Find the locus of a point such that its distance from the point $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module