Question

PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

Answer 1

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 16
⇒ PR + PR = 16
⇒ PR = 8
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 10² = OR² + (8)²
⇒ OR² = 36
⇒ OR = 6
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (8)²
⇒ x² = y² + 64 .....(1)
Again, in right triangle TPO
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y = x² + 64 .....(2)
Solving (1) and (2), we get
y= 10.67

From (1)

$x^2=(10.67{)}^2+64\Rightarrow x^2=113.85+64\Rightarrow x^2=177.85\Rightarrow x=13.34$

∴ TP = 13.34cm