Let C be the circle with centre (0,0) and rad | vedclass

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

Question

(D) Let $AB$ be a chord of the circle $C$ and $M(h, k)$ be the midpoint of $AB$.Since the chord $AB$ subtends an angle of $\frac{2\pi}{3}$ at the cent

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $AB$ be a chord of the circle $C$ and $M(h, k)$ be the midpoint of $AB$.Since the chord $AB$ subtends an angle of $\frac{2\pi}{3}$ at the center $O(0,0)$,the triangle $	riangle OAB$ is an isosceles triangle with $OA = OB = 3$.The line segment $OM$ is the perpendicular bisector of the chord $AB$.In the right-angled triangle $	riangle OAM$,the angle $\angle AOM = \frac{1}{2} \angle AOB = \frac{1}{2} 	imes \frac{2\pi}{3} = \frac{\pi}{3} = 60^{\circ}$.In $	riangle OAM$,we have $OM = OA \cos(60^{\circ})$.Since $M(h, k)$ is the midpoint,the distance $OM = \sqrt{h^2 + k^2}$.Thus,$\sqrt{h^2 + k^2} = 3 \cos(60^{\circ}) = 3 	imes \frac{1}{2} = \frac{3}{2}$.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 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 center is:

Similar Questions

Ferredoxin is a structural component of which of the following?

If the vectors $\overline{AB}=3 \hat{i}+4 \hat{k}$ and $\overline{AC}=5 \hat{i}-2 \hat{j}+4 \hat{k}$ are the sides of the triangle $ABC$,then the length of the median through $A$ is

$A$ wire can be broken by applying a load of $20\,kg$ weight. The force required to break the wire of twice the diameter is ......... $kg\,wt$.

The orange coloured compound formed when $H_2O_2$ is added to $TiO_2$ solution acidified with conc $H_2SO_4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module