In the figure, $∠ A C B = 90^{o}$ and radius of big circle $= 2 c m$ , then the radius of small circle is (in $c m$ ):

3 - 2 \sqrt{2}

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4 - 2 \sqrt{2}

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7 - 4 \sqrt{2}

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6 - 4 \sqrt{2}

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Solution

The correct option is D $6 - 4 \sqrt{2}$
In the diagram, we join the centers and draw perpendiculars form center to $A C$ and $B C .$
In triangle $△ P O Q P O^{2} = P Q^{2} + O Q^{2} {(2 + x)}^{2} = {(2 - x)}^{2} + {(2 - x)}^{2} x^{2} + 4 + 4 x = x^{2} + x^{2} + 4 + 4 - 4 x - 4 x x^{2} + 4 + 4 x = 2 x^{2} + 8 - 8 x x^{2} - 12 x + 4 = 0 x = \frac{12 \pm \sqrt{144 - 16}}{2} = \frac{12 \pm \sqrt{128}}{2} x = 6 - 4 \sqrt{2}$

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In the figure, ∠ACB=90o and radius of big circle =2cm, then the radius of small circle is (in cm):

The correct option is D 6−4√2In the diagram, we join the centers and draw perpendiculars form center to AC and BC. In triangle △POQPO2=PQ2+OQ2(2+x)2=(2−x)2+(2−x)2x2+4+4x=x2+x2+4+4−4x−4xx2+4+4x=2x2+8−8xx2−12x+4=0x=12±√144−162=12±√1282x=6−4√2

In the figure, $∠ A C B = 90^{o}$ and radius of big circle $= 2 c m$ , then the radius of small circle is (in $c m$ ):