overrightarrow{ PA } and overrightarrow{ PB } | vedclass

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(C) In the given circle,$\overline{ OB }$ is the radius and $\overline{ PB }$ is the tangent at point $B$.Since the radius is perpendicular to the tan

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) In the given circle,$\overline{ OB }$ is the radius and $\overline{ PB }$ is the tangent at point $B$.Since the radius is perpendicular to the tangent at the point of contact,$\angle OBP = 90^{\circ}$.Thus,$\Delta POB$ is a right-angled triangle with hypotenuse $OP$.By the Pythagorean theorem:$OP^{2} = OB^{2} + PB^{2}$$13^{2} = 5^{2} + PB^{2}$$169 = 25 + PB^{2}$$PB^{2} = 169 - 25 = 144$$PB = \sqrt{144} = 12$Therefore,$PB = 12$.

$\overrightarrow{ PA }$ and $\overrightarrow{ PB }$ are tangents to $\odot( O , 5) .$ If $OP = 13,$ then $PB = \ldots \ldots \ldots \ldots$

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