A line segment AB of length lambda moves such | vedclass

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

Question

(D) Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$. Since $A$ and $B$ lie on a circle of radius $\lambda$ centered at the origin,$

Vedclass · Accepted Answer

$\frac{\sqrt{19}}{5} \lambda$

Vedclass · Answer

(D) Let the coordinates of $A$ be $(x_1, y_1)$ and $B$ be $(x_2, y_2)$. Since $A$ and $B$ lie on a circle of radius $\lambda$ centered at the origin,$x_1^2 + y_1^2 = \lambda^2$ and $x_2^2 + y_2^2 = \lambda^2$. Given the length $AB = \lambda$,the distance formula gives $(x_2 - x_1)^2 + (y_2 - y_1)^2 = \lambda^2$,which simplifies to $x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2(x_1x_2 + y_1y_2) = \lambda^2$. Substituting the circle equations,we get $2\lambda^2 - 2(x_1x_2 + y_1y_2) = \lambda^2$,so $x_1x_2 + y_1y_2 = \frac{\lambda^2}{2}$. Let $P(h, k)$ be the point dividing $AB$ in the ratio $2:3$. By the section formula,$h = \frac{2x_2 + 3x_1}{5}$ and $k = \frac{2y_2 + 3y_1}{5}$. Then $25(h^2 + k^2) = (2x_2 + 3x_1)^2 + (2y_2 + 3y_1)^2 = 4(x_2^2 + y_2^2) + 9(x_1^2 + y_1^2) + 12(x_1x_2 + y_1y_2)$. Substituting the known values: $25(h^2 + k^2) = 4\lambda^2 + 9\lambda^2 + 12(\frac{\lambda^2}{2}) = 13\lambda^2 + 6\lambda^2 = 19\lambda^2$. Thus,$h^2 + k^2 = \frac{19}{25}\lambda^2$,which represents a circle of radius $\frac{\sqrt{19}}{5}\lambda$.

$A$ line segment $AB$ of length $\lambda$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $\lambda$. Then the locus of the point,that divides the line segment $AB$ in the ratio $2:3$,is a circle of radius

Similar Questions

Let $A = (a, 0)$ and $B = (-a, 0)$ be two fixed points. For $a \in (-\infty, 0)$,point $P$ moves in the plane such that $PA = nPB$ $(n \neq 0, n \neq 1)$. If $n > 1$,then which of the following is true regarding the positions of $A$ and $B$ with respect to the locus of $P$?

$A$ circle passes through the point $(3, 4)$ and cuts the circle $x^2 + y^2 = a^2$ orthogonally. The locus of its centre is a straight line. If the distance of this straight line from the origin is $25$,then $a^2$ is equal to:

Find the locus of point $P$ such that $2PA = 3PB$,where $A = (0, 0)$ and $B = (4, -3)$.

$A$ circle passing through the origin cuts the coordinate axes at $A$ and $B$. If the straight line $AB$ passes through a fixed point $(x_1, y_1)$,then the locus of the centre of the circle is:

If $P$ is a variable point such that the sum of the distances from $P$ to the points $A(2,2)$ and $B(2,-2)$ is $4$,then the locus of $P$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module