Let C be the circle with centre at (1, 1) and | vedclass

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(B) Let the radius of circle $T$ be $r$. Since $T$ is centered at $(0, y)$ and passes through the origin $(0, 0)$,its radius $r$ is the distance betwe

Vedclass · Accepted Answer

$\frac{1}{4}$

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(B) Let the radius of circle $T$ be $r$. Since $T$ is centered at $(0, y)$ and passes through the origin $(0, 0)$,its radius $r$ is the distance between $(0, y)$ and $(0, 0)$,so $r = |y|$.Given that $T$ touches circle $C$ externally,the distance between their centers must be equal to the sum of their radii.The center of $C$ is $(1, 1)$ and its radius is $R = 1$.The center of $T$ is $(0, y)$ and its radius is $r = y$ (assuming $y > 0$).The distance between centers $(1, 1)$ and $(0, y)$ is $\sqrt{(1-0)^2 + (1-y)^2} = \sqrt{1 + (1-y)^2}$.Equating this to the sum of radii $R + r = 1 + y$,we get:$\sqrt{1 + (1-y)^2} = 1 + y$Squaring both sides:$1 + (1 - 2y + y^2) = (1 + y)^2$$2 - 2y + y^2 = 1 + 2y + y^2$$2 - 2y = 1 + 2y$$4y = 1$$y = \frac{1}{4}$Thus,the radius of $T$ is $\frac{1}{4}$.

Let $C$ be the circle with centre at $(1, 1)$ and radius $= 1$. If $T$ is the circle centred at $(0, y)$,passing through the origin and touching the circle $C$ externally,then the radius of $T$ is equal to:

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