Question

Diameter AB of circle with centre O is extended to C and from C a line is drawn tangent to the circle at P. PT is drawn perpendicular to AB at T. Prove that
$CA\cdot CB-TA\cdot TB=C{T}^2$

Answer 1

Solution:

To prove:

$CA.CB-TA.TB=C{T}^2$

Proof:

$CA.CB=P{C}^2$ (Tangent secant theorem) $..............\left(i\right)$

$OA=OB=OP$

In $△OTP$

$O{P}^2=T{P}^2+O{T}^2$ (Pythagorus theorem)

or, $O{A}^2=T{P}^2+(OA-TB{)}^2$

or, $T{P}^2=O{A}^2-O{A}^2-T{B}^2+2TB.OA$

or, $T{P}^2=TB(2OA-TB)$ or,

$T{P}^2=TB.TA$ $................\left(ii\right)$

In $△PTC$

$P{C}^2=T{P}^2+C{T}^2$ (Pythagorus theorem)

or, $C{T}^2=CA.CB-TB.TA$ ( From $eqn.\left(i\right)$ and $eqn.\left(ii\right)$)

Proved.