Let C be a circle with radius 3 and center at | vedclass

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

Question

(C) Let $AB$ be a chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center $C(0, 0)$.Let $D$ be the midpoint of the chord $AB$. Th

Vedclass · Accepted Answer

$x^2 + y^2 = \frac{9}{4}$

Vedclass · Answer

(C) Let $AB$ be a chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center $C(0, 0)$.Let $D$ be the midpoint of the chord $AB$. Then $CD \perp AB$.In $\Delta ACD$,the angle $\angle ACD = \frac{1}{2} \angle ACB = \frac{1}{2} 	imes \frac{2\pi}{3} = \frac{\pi}{3}$.The radius $CA = 3$.Let the coordinates of the midpoint $D$ be $(h, k)$. Then $CD = \sqrt{h^2 + k^2}$.In the right-angled triangle $\Delta ACD$,we have $\cos(\angle ACD) = \frac{CD}{CA}$.$\cos(\frac{\pi}{3}) = \frac{\sqrt{h^2 + k^2}}{3}$.$\frac{1}{2} = \frac{\sqrt{h^2 + k^2}}{3}$.$\sqrt{h^2 + k^2} = \frac{3}{2}$.Squaring both sides,we get $h^2 + k^2 = \frac{9}{4}$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 = \frac{9}{4}$.

Let $C$ be a circle with radius $3$ and center at $(0, 0)$. The equation of the locus of the midpoint of the chord of the circle that subtends an angle of $\frac{2\pi}{3}$ at the center is:

Similar Questions

$A$ circle is drawn to cut a chord of length $2a$ units along the $X$-axis and to touch the $Y$-axis. The locus of the centre of the circle is

The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$, which subtend a right angle at the origin, is

The locus of the centres of the circles,which touch the circle $x^2 + y^2 = 1$ externally,also touch the $y$-axis and lie in the first quadrant is

The locus of a point which is at a distance of $4$ units from $(3, -2)$ in the $xy$-plane is

The locus of the point of intersection of the tangents to the circle $x^2+y^2=a^2$ which make complementary angles with the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module