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
An isosceles right-angled triangle (a \( 45^{\circ}-45^{\circ}-90^{\circ} \) triangle) always has its sides in the ratio \( 1 : 1 : \sqrt{2} \).
Since \( \angle OTP = 90^{\circ} \) and \( \angle POT = 45^{\circ} \), the remaining angle \( \angle OPT \) is also \( 45^{\circ} \).
Therefore, the two perpendicular sides are equal: \( OT = PT = r \).
The hypotenuse \( OP \) is always \( \sqrt{2} \) times the length of the equal sides: \( OP = r\sqrt{2} \).
This geometric rule lets you find the answer instantly without any trigonometry.
Step 1: Understand the Question:
This question is from Circles.
$PT$ is a tangent to a circle with centre $O$ and radius $r$, touching the circle at $T$. We are given $\angle POT = 45^{\circ}$ and need to find the length $OP$.
Step 2: Note the two angles we already know in triangle $OTP$.
A tangent is always perpendicular to the radius drawn to the point of contact, so
\[ \angle OTP = 90^{\circ} \]
We are given
\[ \angle POT = 45^{\circ} \]
Step 3: Find the third angle and identify the triangle type.
The three angles of triangle $OTP$ add up to $180^{\circ}$, so the remaining angle at $P$ is
\[ \angle OPT = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ} \]
Since $\angle POT = \angle OPT = 45^{\circ}$, the sides facing these equal angles must also be equal, so
\[ OT = PT = r \]
This makes $OTP$ an isosceles right triangle, right-angled at $T$.
Step 4: Apply the Pythagorean theorem to find the hypotenuse $OP$.
In this right triangle, $OP$ is the hypotenuse (it is opposite the right angle at $T$), so
\[ OP^2 = OT^2 + PT^2 \]
Put in $OT = PT = r$:
\[ OP^2 = r^2 + r^2 = 2r^2 \]
Take the square root of both sides:
\[ OP = \sqrt{2r^2} = r\sqrt{2} \]
Final Answer:
The length of $OP$ is $r\sqrt{2}$, which matches option (A).
\[ \boxed{OP = r\sqrt{2}} \]
