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if pab is asecant to circle intersecting the circle at a and b abd pt is atangent segment then prove that pa.pb=pt^2

Posted by Himanshu Himanshu 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

PA .PB = (PN - AN)(PN + BN)
= (PN - AN) (PN + AN) {tex}\left[ \begin{array} { c } { \because O N \perp A B } \\ { \therefore N \text { is the mid-point of } A B } \\ { \Rightarrow A N = B N } \end{array} \right]{/tex}
= PN² - AN²
Applying Pythagoras theorem in right triangle PNO, we obtain
OP² = ON² + PN²
{tex}\Rightarrow{/tex}PN²= OP² - ON²
^{{tex}\therefore{/tex}} PN² - AN²= (OP² - ON²) - AN²
= OP²- (ON² + AN²)
= Op² _ OA ²[Using Pythagoras theorem in {tex}\Delta O N A{/tex}]
= OP²- OT² {tex}[ \because O A = O T = \text { radius } ]{/tex}
From (i) and (ii), we obtain
PA.PB = PN²- AN²and PN² - AN² = OP² - OT²
{tex}\Rightarrow{/tex} PA .PB = OP² - OT²
Applying Pythagoras theorem in {tex}\triangle O T P{/tex}, we obtain
OP²= OT² + PT²
{tex}\Rightarrow{/tex} OP²- OT² = PT²
Thus, we obtain
PA.PB = OP² - OT²
and OP² - OT² = PT²
Hence, PA.PB = PT².

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