Question

PQ is a chord of length 4.8 cm of a circle of radius 3cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Answer 1

Let TR = y and TP = x

We know that the perpendicular drawn from the center to me chord bisects It.

∴ PR = RQ

Now, PR + RQ = 4.8

\(\Rightarrow\) PR + PR = 4.8

\(\Rightarrow\) PR = 2.4

Now, in right triangle POR 

By Using Pythagoras theorem, we have

PO² = OR² + PR²

\(\Rightarrow\) 3² = OR² + (2.4)²

\(\Rightarrow\) OR² = 3.24

\(\Rightarrow\) OR = 1.8

Now, in right triangle TPR 

By Using Pythagoras theorem, we have

TP² = TR² + PR²

\(\Rightarrow\) x² = y² + (2.4)²

\(\Rightarrow\) x² = y² + 5.76    ......(1)

Again, In right triangle TPQ 

By Using Pythagoras theorem, we have

TO² = TP² + PO²

\(\Rightarrow\) (y + 1.8)² = x² + 3²

\(\Rightarrow\) y² + 3.6y + 3.24 = x² + 9

\(\Rightarrow\) y² + 3.6y = x² + 5.76   ......(2)

Solving (1) and (2), we get

x = 4 cm and y = 3.2 cm

∴ TP = 4 cm