Question

$PQ$ is a chord of length $4\ \text{cm}$ of a circle of radius $2.5\ \text{cm}$. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length of $TP$.
 

Hint:

For circle tangent problems, combine the perpendicular from the centre to a chord with the pole–polar relation and apply power of a point.

Answer 1

 

Step 1: Midpoint and perpendicular from the centre.
Let $O$ be the centre of the circle, and $M$ the midpoint of chord $PQ$. Then $OM \perp PQ$.
Given $OP = 2.5\ \text{cm},\ PQ = 4\ \text{cm} $\Rightarrow$ PM = 2\ \text{cm}$.
Thus, $OM = \sqrt{OP^2 - PM^2} = \sqrt{2.5^2 - 2^2} = \sqrt{6.25 - 4} = 1.5\ \text{cm}$.
 

Step 2: Relation between polar and pole.
Since $T$ is the intersection of tangents at $P$ and $Q$, chord $PQ$ is the polar of $T$. For a circle, if the perpendicular distance from $O$ to the polar is $d$, and $OT$ is the distance of the pole from $O$, then: \[ d = \frac{r^2}{OT}. \] Here $d=1.5,\ r=2.5 $$\Rightarrow$ \[ OT=\frac{2.5^2}{1.5}=\frac{6.25}{1.5}=\frac{25}{6}\ \text{cm} \]

Step 3: Tangent length.
By power of a point: $TP^2 = OT^2 - r^2$.
\[ TP = \sqrt{\left(\frac{25}{6}\right)^2 - (2.5)^2} = \sqrt{\frac{625}{36} - \frac{25}{4}} = \sqrt{\frac{400}{36}} = \frac{10}{3}\ \text{cm}. \] \[ \boxed{TP = \tfrac{10}{3}\ \text{cm}} \]