A tangent $A B$ at a point $A$ of a circle of radius $5$ cm meets a line through the center $O$ at a point $B$ , so that $O B =$ 12 cm. Length $A B$ is

10

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12

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9

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\sqrt{119}

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Solution

The correct option is B $\sqrt{119}$
$A B = \sqrt{12^{2} {- 5}^{2}} = \sqrt{144 - 25} = \sqrt{119}$

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Similar questions

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :

Q. Question 3
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D)

\sqrt{119} c m

A tangent at a point $A$ on the circle meets a line through the centre of the circle at a point $B$ so that $O B = 13 c m$ and $A B = 12 c m$ . Find the radius of the circle.

Q. O is the center of a circle of radius 8 cm. The tangent at a point A on the circle cuts a line through O at B such that AB=5 cm. Find OB.

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A tangent AB at a point A of a circle of radius 5 cm meets a line through the center O at a point B, so that OB= 12 cm. Length AB is

The correct option is B √119AB=√122−52=√144−25=√119

A tangent $A B$ at a point $A$ of a circle of radius $5$ cm meets a line through the center $O$ at a point $B$ , so that $O B =$ 12 cm. Length $A B$ is

The correct option is B $\sqrt{119}$
$A B = \sqrt{12^{2} {- 5}^{2}} = \sqrt{144 - 25} = \sqrt{119}$