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

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Question

$C T^{2}=(1-0)^{2}+(1-y)^{2}$

also $C T=1+y$

$	herefore(1+y)^{2}=1+(1-y)^{2}$

$1^{2}+2 y+y^{2}=1+1^{2}-2 y+y^{2}$

$2 y+2 y=1+1^{2}+y^{2}-

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

$C T^{2}=(1-0)^{2}+(1-y)^{2}$

also $C T=1+y$

$	herefore(1+y)^{2}=1+(1-y)^{2}$

$1^{2}+2 y+y^{2}=1+1^{2}-2 y+y^{2}$

$2 y+2 y=1+1^{2}+y^{2}-1^{2}-y^{2}$

$4 y=1$

$\Longrightarrow y=\frac{1}{4}$

Hence 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 origin and touching the circle $C$ externally, then the radius of $T$ is equal

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