The diameter of the circle,whose center lies | vedclass

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(C) Let the center of the circle be $(\alpha, 2-\alpha)$ as it lies on the line $x+y=2$.Since the circle is in the first quadrant,$\alpha > 0$ and $2-

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$3$

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(C) Let the center of the circle be $(\alpha, 2-\alpha)$ as it lies on the line $x+y=2$.Since the circle is in the first quadrant,$\alpha > 0$ and $2-\alpha > 0$,which implies $0 The circle touches the lines $x=3$ and $y=2$. The radius $r$ is the distance from the center to these lines:$r = |3-\alpha| = |2-(2-\alpha)| = |\alpha|$.Since $0 Equating the two expressions for the radius:$3-\alpha = \alpha$$2\alpha = 3$$\alpha = \frac{3}{2}$.The radius $r = \alpha = \frac{3}{2}$.The diameter of the circle is $2r = 2 	imes \frac{3}{2} = 3$.

The diameter of the circle,whose center lies on the line $x+y=2$ in the first quadrant and which touches both the lines $x=3$ and $y=2$,is

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