Question

AB is a chord of length 9.6cm of a circle with centre O and radius 6 cm. If the tangents at A and B intersect at point P then find the length PA.

☆SOLVE THIS WITH TWO DIFFERENT METHODS☆
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Answer 1

Given:-

• AB = 9.6 cm and radius OA = 6 cm

To Find:-

• length of Line Segment PA.

CONSTRUCTION:-

• Join OP and OA .Let OP and AB Intersects at M

• Let PA = x cm and PM = y cm.

Now, PA=PB

• triangle APB is isosceles and so PO is angle bisector of triangle APB.

∵ two tangents to a circle from an external point are equally inclined to the line segment joining the centre to that point.

Also, OP⊥AB

• and OP bisects AB at M [∵ OP is the right bisector of AB]

∵AM=MB = ½ AB = 4.8 cm

• In right △AMO, we have

OA=6cm

and AM=4.8cm.

OM=

$\sqrt{ {6}^{2} } - {(4.8)}^{2}$

=

$\sqrt{12.96}$

=

$3.6cm$

In right △PAO, we have

AP²= PM²+ AM²

⇒x²=y²+(4.8)²

⇒x²=y²+23.04...(i)

In right △PAO, we have

OP² = PA² + OA²

[Note ∠PAO=90∘, since AO is the radius at the point of contact]

⇒(y+3.6)²

=x²+6²

⇒y²+7.2y+12.96

=x²+36

⇒7.2y=46.08 [using (i)]

⇒y=6.4cm

AND

x²= (6.4)² + 23.04

= 40.96 + 23.04 = 64

⇒x=√64=8

∴ PA=8cm.

Answer 2

OP is the perpendicular Bisector of chord AB

$\therefore$ $\sf{AR=BR=\frac{1}{2}AB}$

$\longrightarrow$$\sf{\frac{1}{2}=9.6=4.8cm}$

$\sf{In\:right\:∆ARO,we\:have}$

$\longrightarrow$$\sf\red{OR = \sqrt{ {or}^{2} - {AR}^{2} }}$

$\longrightarrow$$\sf\red{ \sqrt{ {6}^{2} - {(4.8)}^{2} }}$

$\longrightarrow$$\sf\red{\sqrt{10.8 \times 1.2}}$

$\longrightarrow$$\sf\red{\sqrt{12.96}}$

$\longrightarrow$$\sf\red{3.6cm}$

$\sf\purple{∆ARP~∆ORA}$

$\therefore$ $\sf\blue{\frac{PA}{AR}=\frac{OA}{OR}}$

Hence, $\sf\blue{PA=\frac{OA}{OQ}×AR}$

$\sf{\frac{6×4.8}{3.6}=8cm}$