The locus of the midpoint of the chords of th | vedclass

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

Question

(A) The given circle is $x^2 + y^2 - 2x - 2y - 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -1, c = -2$.The centre $C$ i

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 2y + 1 = 0$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 - 2x - 2y - 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -1, c = -2$.The centre $C$ is $(-g, -f) = (1, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 1 - (-2)} = \sqrt{4} = 2$.Let $M(h, k)$ be the midpoint of a chord $AB$. The triangle $CAM$ is a right-angled triangle at $M$,where $CA = r = 2$ and $\angle ACM = \frac{120^\circ}{2} = 60^\circ = \frac{\pi}{3}$ radians.In $	riangle CAM$,$\cos(\angle ACM) = \frac{CM}{CA}$.$\cos(60^\circ) = \frac{CM}{2}$ $\Rightarrow \frac{1}{2} = \frac{CM}{2}$ $\Rightarrow CM = 1$.The distance $CM$ is the distance between $(h, k)$ and $(1, 1)$,so $CM^2 = (h-1)^2 + (k-1)^2$.Since $CM = 1$,we have $(h-1)^2 + (k-1)^2 = 1^2$.$h^2 - 2h + 1 + k^2 - 2k + 1 = 1$.$h^2 + k^2 - 2h - 2k + 1 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 2x - 2y + 1 = 0$.

The locus of the midpoint of the chords of the circle $x^2 + y^2 - 2x - 2y - 2 = 0$ which makes an angle of $120^\circ$ at the centre 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:

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 $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ 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

$A$ circle passing through the origin cuts the coordinate axes at $A$ and $B$. If the straight line $AB$ passes through a fixed point $(x_1, y_1)$,then the locus of the centre of the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module