Question

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 :

• A. $r\sqrt{2}$
• B. $\sqrt{2r}$
• C. $2r$
• D. $r^2$

Hint:

In a right-angled triangle with an angle of $45^\circ$, the other acute angle is also $45^\circ$ ($90^\circ - 45^\circ = 45^\circ$).
This makes it an isosceles right-angled triangle where the two perpendicular sides are equal: $OT = PT = r$.
Using the Pythagorean theorem directly:
\[ OP = \sqrt{OT^2 + PT^2} = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2} \] This visual and geometric approach bypasses trigonometry completely and is highly intuitive.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question belongs to the topic of "Circles" and "Trigonometry".
We are given a circle with center $O$ and radius $r$.
$PT$ is a tangent to this circle at point $T$, and we are given $\angle POT = 45^\circ$.
We need to determine the length of the line segment $OP$ in terms of the radius $r$.

Step 2: Key Formula or Approach:
We use two main concepts:

• Tangent-Radius perpendicularity: The tangent at any point of a circle is perpendicular to the radius through the point of contact. This means $OT \perp PT$, making $\angle OTP = 90^\circ$.

• Right-Angled Triangle Trigonometry: In the right-angled triangle $\Delta OTP$, we can use trigonometric ratios. Here, $OT$ is the adjacent side to $\angle POT$, and $OP$ is the hypotenuse. We can use the cosine ratio:
\[ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \]

Step 3: Detailed Explanation:

• Identify the given elements from the problem statement and figure:

• Radius $OT = r$

• Angle $\angle POT = 45^\circ$

• Since $PT$ is a tangent to the circle at point $T$, the angle formed between the radius $OT$ and the tangent $PT$ is a right angle:
\[ \angle OTP = 90^\circ \] Therefore, $\Delta OTP$ is a right-angled triangle with the hypotenuse being the segment $OP$.

• In the right-angled triangle $\Delta OTP$, apply the cosine trigonometric ratio for $\angle POT = 45^\circ$:
\[ \cos(\angle POT) = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{OT}{OP} \]

• Substitute the known values into the equation:
\[ \cos(45^\circ) = \frac{r}{OP} \]

• Use the standard value of $\cos(45^\circ) = \frac{1}{\sqrt{2}}$:
\[ \frac{1}{\sqrt{2}} = \frac{r}{OP} \]

• Rearrange the equation to solve for $OP$:
\[ OP = r\sqrt{2} \]

Step 4: Final Answer:
The length of the segment $OP$ is $r\sqrt{2}$, which corresponds to Option (A).