The length of the chord joining points (4 cos | vedclass

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Question

(A) The equation of the circle is $x^2+y^2=16$,which represents a circle centered at the origin $(0,0)$ with radius $r=4$.Let the two points be $A = (

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2=16$,which represents a circle centered at the origin $(0,0)$ with radius $r=4$.Let the two points be $A = (4 \cos 	heta, 4 \sin 	heta)$ and $B = (4 \cos (	heta+60^{\circ}), 4 \sin (	heta+60^{\circ}))$.These points represent points on the circle with polar angles $	heta$ and $	heta+60^{\circ}$ respectively.The angle subtended by the chord $AB$ at the center $O(0,0)$ is $\Delta \phi = (	heta+60^{\circ}) - 	heta = 60^{\circ}$.Since $OA = OB = r = 4$ and the included angle $\angle AOB = 60^{\circ}$,the triangle $	riangle OAB$ is an equilateral triangle.Therefore,the length of the chord $AB$ is equal to the radius of the circle.$AB = r = 4$.

The length of the chord joining points $(4 \cos \theta, 4 \sin \theta)$ and $(4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$ on the circle $x^2+y^2=16$ is

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