Question

If a tangent PT and a line segment PBA is drawn to a circle with centre O. If OL is perpendicular to AB, prove that PA. PB=PT^2.

Answer 1

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Answer 2

$\underline { \pink { Given : }}$

A tangent PT and a line segment PBA is drawn to a circle with center 'O' .

$OL \perp AB$

$\underline { \pink { To \:prove : }}$

$\red { PA \times PB } \green {= PT^{2}}$

$\underline { \pink { Construction : }}$

$Join \: OP , OT \: and \: OA$

$\underline { \green { Proof : }}$

$The\:foot \;of \:the \: perpendicular \:from \\the\:centre\:to \:a \:chord \: bisects \:the \:chord.$

$OL \perp AB \implies AL = LB \: --(1)$

$Now, \red {PA \times PB} \\= ( PL - AL )(PL+LB) \\=( PL - AL )(PL+AL)\: [From \:(1) ]\\= PL^{2} - AL^{2} \\= ( OP^{2} - OL^{2} ) - AL^{2}$

$\blue {( In \:right \:angle {OLP}, we \:have }\\\blue { OP^{2} = OL^{2} + PL^{2} }$

$\blue { PL^{2} = OP^{2} - OL^{2} ) }$

$= OP^{2} - ( OL^{2} + AL^{2} )$

$= OP^{2} - OA^{2}$

$\blue {( In \:right \:angle {OLA}, we \:have }\\\blue {OA^{2} = OL^{2} + AL^{2} )}$

$= OP^{2} - OA^{2} \: \blue { ( OA = OT = r)}$

$= PT^{2}$

$\blue { ( \angle {OTP } = 90\degree }\\\blue {OP^{2} = OT^{2} + PT^{2} \implies OP^{2} - OT^{2} = PT^{2} )}$

$Hence , \: PA \times PB = PT^{2}$

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