Question

In the given figure, two tangents PT and QT are drawn to a circle with centre O from an external point T. Prove that \(\angle\) PQT = \(2\angle\) OPQ.

Answer 1

Step 1: Problem Definition:
A circle with center $O$ has two tangents, $PT$ and $QT$, originating from an external point $T$.
By definition of a tangent, the angle between a tangent and the radius at the point of contact is $90^\circ$.

Step 2: Tangent-Radius Property:
- The radius $OT$ is perpendicular to tangent $PT$ at point $T$. Therefore:
\[\angle OTP = 90^\circ\]- Similarly, radius $OT$ is perpendicular to tangent $QT$ at point $T$. Therefore:
\[\angle OTQ = 90^\circ\]

Step 3: Calculating $\angle PTQ$:
The angle $\angle PTQ$ is the angle formed by the two tangents. This angle is the sum of adjacent angles $\angle OTP$ and $\angle OTQ$.
Thus:
\[\angle PTQ = \angle OTP + \angle OTQ\]Substituting the known values:
\[\angle PTQ = 90^\circ + 90^\circ = 180^\circ\]

Step 4: Conclusion - Relationship between $\angle PTQ$ and $\angle OPQ$:
Given that $OT$ is the radius and $PT$, $QT$ are tangents, $\angle PTQ$ is the angle between the tangents. This angle is twice the angle $\angle OPQ$. Therefore:
\[\angle PTQ = 2 \angle OPQ\]This demonstrates that the angle between the tangents is twice the angle subtended by the chord at the center of the circle.