AC is the tangent to the circle having O as the centre at point B. What would be the length of AM?

3 cm

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2 cm

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2.5 cm

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None of these

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Solution

The correct option is B 2 cm

As we know,
$O B ⟂ A C$ ,
So, $∠ O B A = 90^{0} .$
Apply Pythagoras Theorem in $Δ O B A$
$O A^{2} = O B^{2} + A B^{2}$
$O A^{2} = (3)^{2} + (4)^{2}$
$O A^{2} = 9 + 16 = 25$
$O A = \sqrt{25}$
$O A = 5 c m$
And we know that,
$O M = O B = 3 c m$ [ Equal radii of the circle]
So,
$A M = O A - O M$
$A M = 5 - 3 = 2 c m$
Hence, b is the correct option.

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AC is the tangent to the circle having O as the centre at point B. What would be the length of AM?

The correct option is B 2 cm As we know, OB⟂AC, So, ∠OBA=900. Apply Pythagoras Theorem in ΔOBA OA2=OB2+AB2 OA2=(3)2+(4)2 OA2=9+16=25 OA=√25 OA=5cm And we know that, OM=OB=3cm [ Equal radii of the circle] So, AM=OA−OM AM=5−3=2cm Hence, b is the correct option.