The locus of the centre of the circles passin | vedclass

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

Question

(C) Let the centre of the circle be $(h, k)$.Since the circle passes through the origin $(0, 0)$,its radius $R$ is given by $R^2 = h^2 + k^2$.The equa

Vedclass · Accepted Answer

a parabola

Vedclass · Answer

(C) Let the centre of the circle be $(h, k)$.Since the circle passes through the origin $(0, 0)$,its radius $R$ is given by $R^2 = h^2 + k^2$.The equation of the circle is $(x-h)^2 + (y-k)^2 = h^2 + k^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky = 0$.The circle cuts a chord of length $2$ units on the line $x=1$. The length of the chord is given by $2\sqrt{R^2 - d^2} = 2$,where $d$ is the perpendicular distance from the centre $(h, k)$ to the line $x=1$.Thus,$\sqrt{R^2 - d^2} = 1$,or $R^2 - d^2 = 1$.Here,$R^2 = h^2 + k^2$ and $d = |h-1|$.Substituting these values,we get $(h^2 + k^2) - (h-1)^2 = 1$.$h^2 + k^2 - (h^2 - 2h + 1) = 1$$h^2 + k^2 - h^2 + 2h - 1 = 1$$k^2 + 2h - 1 = 1$$k^2 = 2 - 2h$Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2(1-x)$,which represents a parabola.

The locus of the centre of the circles passing through the origin and cutting off a chord of length $2$ units on the line $x=1$ is

Similar Questions

The image of every point lying on the curve $x^2+y^2=1$ in the line $x+y=1$ satisfies the equation:

The locus of the midpoints of the chords of the circle $x^2 + y^2 = a^2$ which are parallel to $y = 2x$ is:

$A$ point $P(x, y)$ divides the line segment joining the points $(5, 0)$ and $(10 \cos \theta, 10 \sin \theta)$ in the ratio $2 : 3$. Find the locus of point $P$ as $\theta$ varies.

$A$ circle cuts a chord of length $4a$ on the $x$-axis and passes through a point on the $y$-axis,distant $2b$ from the origin. Then the locus of the center of this circle is

The point of concurrence of all the chords of the curve $3x^2 - y^2 - 2x + 4y = 0$ which subtend a right angle at the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module