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

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

Question

(A) The points are $P = (4 \cos 	heta, 4 \sin 	heta)$ and $Q = (4 \cos (	heta+60^{\circ}), 4 \sin (	heta+60^{\circ}))$.Using the distance formula,

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The points are $P = (4 \cos 	heta, 4 \sin 	heta)$ and $Q = (4 \cos (	heta+60^{\circ}), 4 \sin (	heta+60^{\circ}))$.Using the distance formula,the length of the chord $PQ$ is given by:$PQ = \sqrt{(4 \cos (	heta+60^{\circ}) - 4 \cos 	heta)^2 + (4 \sin (	heta+60^{\circ}) - 4 \sin 	heta)^2}$$PQ = 4 \sqrt{(\cos (	heta+60^{\circ}) - \cos 	heta)^2 + (\sin (	heta+60^{\circ}) - \sin 	heta)^2}$$PQ = 4 \sqrt{\cos^2 (	heta+60^{\circ}) + \cos^2 	heta - 2 \cos (	heta+60^{\circ}) \cos 	heta + \sin^2 (	heta+60^{\circ}) + \sin^2 	heta - 2 \sin (	heta+60^{\circ}) \sin 	heta}$Using $\sin^2 A + \cos^2 A = 1$:$PQ = 4 \sqrt{1 + 1 - 2 (\cos (	heta+60^{\circ}) \cos 	heta + \sin (	heta+60^{\circ}) \sin 	heta)}$Using $\cos (A-B) = \cos A \cos B + \sin A \sin B$:$PQ = 4 \sqrt{2 - 2 \cos ((	heta+60^{\circ}) - 	heta)}$$PQ = 4 \sqrt{2 - 2 \cos 60^{\circ}}$Since $\cos 60^{\circ} = \frac{1}{2}$:$PQ = 4 \sqrt{2 - 2 	imes \frac{1}{2}} = 4 \sqrt{2 - 1} = 4 	imes 1 = 4$.

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

Similar Questions

The area (in sq. units) of the region bounded by the circle $x^2+y^2=64$, positive $x$-axis and the line $y=\sqrt{3}x$ is

The number of common tangents to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6x-8y-24=0$ is

If a point $P(\alpha, \beta)$ on the line $y=1$ is such that the two distinct chords drawn on $x^2+y^2-\alpha x-y=0$ from $P$ are bisected by the $x$-axis,then

The area of an equilateral triangle inscribed in the circle $x^2+y^2-6x+2y-28=0$ is . . . . . . sq. units.

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module