The locus of the mid-points of the chords of | vedclass

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

Question

(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the centre $C = (1, 2)$ and rad

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 4y - 7 = 0$

Vedclass · Answer

(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the centre $C = (1, 2)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 2^2 - (-11)} = \sqrt{1 + 4 + 11} = \sqrt{16} = 4$.Let $(h, k)$ be the mid-point of a chord. The distance $d$ from the centre $(1, 2)$ to the chord is given by $d = \sqrt{(h-1)^2 + (k-2)^2}$.In the right-angled triangle formed by the centre,the mid-point,and an endpoint of the chord,the angle at the centre is $60^o / 2 = 30^o$.Thus,$\cos 30^o = \frac{d}{r} = \frac{\sqrt{(h-1)^2 + (k-2)^2}}{4}$.Since $\cos 30^o = \frac{\sqrt{3}}{2}$,we have $\frac{\sqrt{(h-1)^2 + (k-2)^2}}{4} = \frac{\sqrt{3}}{2}$.Squaring both sides,$\frac{(h-1)^2 + (k-2)^2}{16} = \frac{3}{4}$.$(h-1)^2 + (k-2)^2 = 12$.$h^2 - 2h + 1 + k^2 - 4k + 4 = 12$.$h^2 + k^2 - 2h - 4k - 7 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - 2x - 4y - 7 = 0$.

The locus of the mid-points of the chords of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ which subtend $60^o$ at the centre is

Similar Questions

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 $t$ is a parameter,$A = (a \sec t, b \tan t)$,$B = (-a \tan t, b \sec t)$,and $O = (0, 0)$,then the locus of the centroid of $\triangle OAB$ is:

Find the locus of the midpoint of a chord of the circle $x^2 + y^2 = a^2$ which subtends a right angle at the center.

The locus of a point which moves such that the sum of the squares of its distances from the three vertices of a triangle is constant,is a circle whose centre is at the

Tangents are drawn from the point $(17,7)$ to the circle $x^2+y^2=169$.
$STATEMENT-1$: The tangents are mutually perpendicular.
$STATEMENT-2$: The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^2+y^2=338$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module