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(A) Let the circle be $x^2 + y^2 = 5^2$. The line $AB$ divides the circumference in ratio $1:2$,so the arc $AB$ subtends an angle of $\frac{1}{3} 	im

Vedclass · Accepted Answer

$x^2 + y^2 = \frac{175}{4}$

Vedclass · Answer

(A) Let the circle be $x^2 + y^2 = 5^2$. The line $AB$ divides the circumference in ratio $1:2$,so the arc $AB$ subtends an angle of $\frac{1}{3} 	imes 360^{\circ} = 120^{\circ}$ at the center $O$.Thus,$\angle AOB = 120^{\circ}$.Let $M$ be the midpoint of $AB$. Then $\angle AOM = 60^{\circ}$.In $	riangle OAM$,$OM = 5 \cos 60^{\circ} = \frac{5}{2}$ and $AM = 5 \sin 60^{\circ} = \frac{5\sqrt{3}}{2}$.Since $ABCD$ is a rectangle,the height of the rectangle is the distance between the chord $AB$ and the tangent $CD$.The distance from the center $O$ to the chord $AB$ is $OM = \frac{5}{2}$.The distance from the center $O$ to the tangent $CD$ is the radius $R = 5$.Thus,the height of the rectangle $AD = BC = 5 - \frac{5}{2} = \frac{5}{2}$.Let $C = (x, y)$. Since $CD$ is horizontal,$y$ coordinate of $C$ is $5$.However,considering the locus of points $C$ and $D$ as the line $AB$ varies,let the distance of the chord from the center be $d = \frac{5}{2}$.The points $C$ and $D$ lie on a line parallel to $AB$ at distance $5$ from the center.Let the coordinates of $C$ be $(x, y)$. The distance of $C$ from the origin is $\sqrt{x^2 + y^2}$.From the geometry,the distance of $C$ from the center is $OC = \sqrt{ON^2 + NC^2}$.Here $ON = 5$ (distance to tangent) and $NC = \frac{AB}{2} = \frac{5\sqrt{3}}{2}$.$OC^2 = 5^2 + (\frac{5\sqrt{3}}{2})^2 = 25 + \frac{75}{4} = \frac{100+75}{4} = \frac{175}{4}$.Thus,the locus is $x^2 + y^2 = \frac{175}{4}$.

$A$ variable straight line $AB$ divides the circumference of the circle $x^2 + y^2 = 25$ in the ratio $1 : 2$. If a tangent $CD$ is drawn to the smaller arc parallel to $AB$,such that $ABCD$ is a rectangle,then the locus of $C$ and $D$ is:

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