PQ is a chord of length 9cm of a circle of radius 6 cm. The tangents to the circle at points P and Q intersect each other at an external point T. Find the length of tangent TP.

Step 1: Understand the Geometry

We are given a circle with center O and radius $OP = OQ = 6$ cm. A chord $PQ$ has a length of $9$ cm. Tangents from an external point $T$ touch the circle at $P$ and $Q$. We need to find the length of the tangent $TP$.

Step 2: Identify Key Properties and Triangles

From the properties of tangents, we know that tangents drawn from an external point to a circle are equal in length, so $TP = TQ$. Also, the radius is perpendicular to the tangent at the point of contact, so $\angle OPT = 90^\circ$. This means $\triangle OPT$ is a right-angled triangle.

Step 3: Find the area of triangle OPQ using Heron's formula

First, we find the semi-perimeter $s$ of $\triangle OPQ$. Then, we use Heron's formula to calculate the area of $\triangle OPQ$. This area will be useful in finding the altitude from O to PQ.

Step 4: Find the altitude from O to PQ

Let $OM$ be the altitude from $O$ to $PQ$. We can also express the area of $\triangle OPQ$ as $\frac{1}{2} \times \text{base} \times \text{height}$. Using the calculated area, we can find the length of $OM$.

Step 5: Find PM using Pythagorean theorem in triangle OMP

The perpendicular from the center to a chord bisects the chord, so $PM = \frac{1}{2} PQ = \frac{9}{2}$ cm. We can verify this using the Pythagorean theorem in $\triangle OMP$. This step confirms our calculations for $OM$ and $PM$.

Step 6: Use similar triangles to find TP

Consider $\triangle OMP$ and $\triangle OPT$. Both are right-angled triangles. $\angle OMP = 90^\circ$ and $\angle OPT = 90^\circ$. Also, $\angle POM$ is common to both triangles. Therefore, $\triangle OMP \sim \triangle OPT$ by AA similarity. Using the ratio of corresponding sides, we can find the length of $PT$.

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