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Maths / Circles / Length of Tangents to a Circle
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## Length of Tangents to a Circle

Length of Tangents to a Circle: it is the distance of the point P to the point of contact on the circle that is point A and Point B. Or in other words PA and PB are the lengths on tangent from point P to the circle

 Theorem: The length of tangents drawn from an external point to a circle are equal.Given :A circle with centre O. PA and PB to tangents from point P.To Prove: PA = PB Proof:  Join OA an OB  In  $\dpi{120} \large \Delta OAP$  and  $\fn_jvn \large \dpi{120} \large \Delta OBP$                OA = OB                          [radii of the same circle]                OP = OP                          [ common ]   $\dpi{120} \large \angle OAP =\angle OBP$   [ each $\dpi{120} \large 90^{\circ}$]Thus $\dpi{120} \large \Delta OAP \cong \Delta OBP$    [ RHS criterion]         PA = PB  [CPCT]Thus, lengths of tangents from an external point to a circle are equal.Corollary: Theline joining the centre of the circle and the external point i.e. OP is the angle bisector of $\fn_cm \angle$ AOB and $\fn_cm \angle$APB.         As    $\dpi{120} \large \Delta OAP\cong \Delta OBP$,         $\dpi{120} \large \angle APO=\angle BPO$    [By CPCT]$\dpi{120} \large \Rightarrow$  $\dpi{120} \large PO$  is the angle bisectors of $\dpi{120} \large \angle APB$ Similarly OP is the angle bisector of $\fn_cm \angle$ AOB

ILLUSTRATION : A quadrilateral ABCD is drawn to circumscribe a circle. Prove that      AB + CD =AD + BC

 Solution    A quadrilateral ABCD circumscribes a cricle. Since the lengths of tangents from an external point to a circle are equal, we get     AP = AS                    [ lengths of tangents from point A ]     BP = BQ                   [ lengths of tangents from point B ]     CR = CQ                  [ lengths of tangents from point C ]     DR = DS                  [ lengths of tangents from point D ]Adding the above expressions, we get      AP + BP + CR + DR = AS + BQ + CQ + DS$\dpi{120} \large \Rightarrow$  ( AP + BP ) + ( CR + DR ) = ( AS + DS ) + ( BQ + CQ )$\dpi{120} \large \Rightarrow$              AB + CD = AD + BCHence Proved

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