Question

$\overline{PT}$ is a tangent of circle with centre O at point T. A secant passing through P in at A and B. If P-B-A, PT$=12$ and PB$=4$, then PA ________.

• A. $16$
• B. $48$
• C. $24$
• D. $36$

Answer 1

Answer: (D)

$36$

Here $PBA$ is recant intersecting the circle with centre $O$ at $A$ and $B$ and a tangent $PT$ at $T$

Draw $OM$ perpendicular to $AB$ and join $OB,OA$ and $OP$

Since $OM$ is perpendicular to $AB$ and a line drawn through center is perpendicular to chord it bisect it also.

$\therefore BM=MA$

here $PB=PA=(PM-BM)+(PM+MA)$

Since $BM=MA$

$\therefore PB\times PA=(PM-BM)\times (PM+BM)$

$\therefore PA\times PB=P{M}^2-B{M}^2\to \left(1\right)$

$\to$ As $△PMO$ is right angled triangle

$\therefore$ By phythagoras theorem,

$P{M}^2=O{P}^2-O{M}^2$

$\to$ Also $△BMO$ is right angle triangle

$\therefore B{M}^2=O{B}^2-O{M}^2$

Substituting back in equation $\left(1\right)$

$PA\times PB=O{P}^2-O{M}^2-(O{B}^2-O{M}^2)$

$\therefore PA\times PB=O{P}^2-O{B}^2$

$\therefore PA\times PB=O{P}^2-O{T}^2$ ($OB=OT$ as it is radius of circle)

$\therefore PA\times PB=P{T}^2$ (By using phythagoras theorem in $△POT$)

Therefore according to given values

$PA\times PB=P{T}^2$

$\therefore PA\times 4=(12{)}^2$

$\therefore PA\times 4=144$

$\therefore PA=36cm$