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(C) The initial center of the circle is $C(5, 5)$ as it touches both axes in the first quadrant with radius $r = 5$.When the circle rolls one complete

Vedclass · Accepted Answer

$(x - 5 - 10\pi )^{2} + (y - 5)^{2} = 25$

Vedclass · Answer

(C) The initial center of the circle is $C(5, 5)$ as it touches both axes in the first quadrant with radius $r = 5$.When the circle rolls one complete revolution along the $x-$axis,the distance covered by the center is equal to the circumference of the circle,which is $2\pi r = 2\pi(5) = 10\pi$.The new center $D$ will have coordinates $(5 + 10\pi, 5)$ while the radius remains $5$.The equation of the circle in the new position is $(x - (5 + 10\pi))^{2} + (y - 5)^{2} = 5^{2}$,which simplifies to $(x - 5 - 10\pi)^{2} + (y - 5)^{2} = 25$.

$A$ circle of radius $5$ units touches the axes in the first quadrant. If the circle rolls along the $x-$axis in the positive $x-$direction for one complete revolution,find its equation in the new position.

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