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

If $A(\cos \alpha, \sin \alpha)$,$B(\sin \alpha, -\cos \alpha)$,and $C(1, 2)$ are the vertices of a $\Delta ABC$,then as $\alpha$ varies,the locus of its centroid is

If a circle of radius $R$ passes through the origin $O$ and intersects the coordinate axes at $A$ and $B,$ then the locus of the foot of the perpendicular from $O$ on $AB$ is

Find the locus of a point such that the sum of the squares of its distances from the axes is $4$.

The angle between the tangents drawn from a point $P$ to the circle $x^2 + y^2 + 4x - 2y - 4 = 0$ is $60^{\circ}$. The locus of $P$ is:

The number of pairs $(a, b)$ of positive real numbers satisfying $a^4+b^4 < 1$ and $a^2+b^2 > 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module