The perimeter of the locus of the point P whi | vedclass

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

Question

(B) Let $Q = (3 \cos 	heta, 3 \sin 	heta)$ be a point on the circle $x^2+y^2=9$.Let $P = (h, k)$ be the point that divides the line segment $QA$ in

Vedclass · Accepted Answer

$4 \pi$

Vedclass · Answer

(B) Let $Q = (3 \cos 	heta, 3 \sin 	heta)$ be a point on the circle $x^2+y^2=9$.Let $P = (h, k)$ be the point that divides the line segment $QA$ in the ratio $1:2$.Using the section formula,the coordinates of $P$ are given by:$h = \frac{1(4) + 2(3 \cos 	heta)}{1+2} = \frac{4 + 6 \cos 	heta}{3}$$k = \frac{1(4) + 2(3 \sin 	heta)}{1+2} = \frac{4 + 6 \sin 	heta}{3}$Rearranging these equations:$3h - 4 = 6 \cos 	heta$$3k - 4 = 6 \sin 	heta$Squaring and adding both equations:$(3h - 4)^2 + (3k - 4)^2 = (6 \cos 	heta)^2 + (6 \sin 	heta)^2$$9(h - \frac{4}{3})^2 + 9(k - \frac{4}{3})^2 = 36$$(h - \frac{4}{3})^2 + (k - \frac{4}{3})^2 = 4$This represents a circle with radius $r = \sqrt{4} = 2$.The perimeter of the locus is $2 \pi r = 2 \pi (2) = 4 \pi$.

The perimeter of the locus of the point $P$ which divides the line segment $QA$ internally in the ratio $1:2$,where $A=(4,4)$ and $Q$ lies on the circle $x^2+y^2=9$ is

Similar Questions

The centres of those circles which touch the circle $x^{2} + y^{2} - 8x - 8y - 4 = 0$ externally and also touch the $x$-axis,lie on:

The point of concurrence of all the chords of the curve $3x^2 - y^2 - 2x + 4y = 0$ which subtend a right angle at the origin is

The locus of the centre of a circle which passes through the origin and cuts off a length of $4$ units from the line $x=3$ is

Without changing the direction of the coordinate axes,the origin is shifted to $(h, k)$ such that the linear (first-degree) terms in the equation $x^2 + y^2 - 4x + 6y - 7 = 0$ are eliminated. Then the point $(h, k)$ is:

The locus of the centre of the circles which touch both the circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4ax$ externally is

Vedclass Test Series

Exam Paper Generator

Online Exam Module