Question : PQ is a chord of length 8 cm of a circle with centre O and radius 5 cm. The tangents at P and Q intersect at a point T. The length of TP is:
Option 1: $\frac{20}{3}\ \text{cm}$
Option 2: $\frac{21}{4}\ \text{cm}$
Option 3: $\frac{10}{3}\ \text{cm}$
Option 4: $\frac{15}{4}\ \text{cm}$
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Correct Answer: $\frac{20}{3}\ \text{cm}$
Solution :
Join OP and OT.
Let OT intersect PQ at a point R.
Then, TP = TQ (The lengths of tangents drawn from an external point to a circle are equal) and $\angle$PTR = $\angle$QTR.
⇒ TR $\perp$ PQ and TR bisect PQ.
⇒ PR = RQ = 4 cm
OP = 5 cm
So, OR
2
= OP
2
– PR
2
= $\sqrt{5^2-4^2}=\sqrt{9}=3$ cm
Let TP = $x$ cm
and TR = $y$ cm
From right ΔTRP, we get,
TP
2
= TR
2
+ PR
2
⇒ $x$
2
= $y$
2
+ 16
⇒ $x$
2
− $y$
2
= 16-------------------------(i)
From right ΔOPT, we get
TP
2
+ OP
2
= OT
2
⇒ $x$
2
+ 5
2
= ($y$ + 3)
2
[$\because$ OT
2
= (OR + RT)
2
]
⇒ $x$
2
− $y$
2
= 6$y$ − 16------------------(ii)
From (i) and (ii), we get,
6$y$ − 16 = 16
⇒ 6$y$ = 32
⇒ $y$ = $\frac{16}{3}$
Putting the value of $y$ in equation (i), we get,
$x$
2
$=16 + (\frac{16}{3})^2=16+\frac{256}{9}=\frac{400}{9}=\frac{20}{3}\ \text{cm}$
Hence, the correct answer is $\frac{20}{3}\ \text{cm}$.
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