In the given figure, PT is a tangent to the circle with centre O and radius r. If \(\angle POT = 45^\circ\), then the length of OP is :
Show Hint
In any right-angled triangle with an angle of \(45^\circ\), the triangle is an isosceles right triangle.
The two legs are equal in length, so \(OT = PT = r\).
By Pythagoras theorem, the hypotenuse is always \(\sqrt{2}\) times the length of the leg, which gives \(OP = r\sqrt{2}\) immediately.
Step 1: Work out all three angles of the triangle first.
Since $PT$ is a tangent at $T$, the radius $OT$ is perpendicular to it, so $\angle OTP = 90^{\circ}$. We are given $\angle POT = 45^{\circ}$. The third angle of the triangle is:
\[ \angle OPT = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ} \]
Step 2: Notice the triangle is isosceles right angled.
Since $\angle POT = \angle OPT = 45^{\circ}$, the sides opposite these equal angles are also equal, so $OT = PT = r$. This means $\Delta OTP$ is an isosceles right triangle with its two legs equal to $r$ and the right angle at $T$.
Step 3: Find the hypotenuse using the Pythagoras theorem instead of a cosine ratio.
$OP$ is the hypotenuse of this right triangle, so:
\[ OP^{2} = OT^{2} + PT^{2} = r^{2} + r^{2} = 2r^{2} \]
\[ OP = \sqrt{2r^{2}} = r\sqrt{2} \]
Final Answer:
The length of $OP$ is $r\sqrt{2}$, matching option (A).
\[ \boxed{OP = r\sqrt{2}} \]
