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 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.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a circle centered at \(O\) with a radius of length \(r\).
\(PT\) is a tangent to the circle at point \(T\).
We need to determine the length of the segment \(OP\) given that \(\angle POT = 45^\circ\).

Step 2: Key Formula or Approach:
By the properties of tangents, a tangent at any point on a circle is perpendicular to the radius through the point of contact.
Therefore, \(\angle OTP = 90^\circ\), making \(\Delta OTP\) a right-angled triangle.
We can use basic right-angle trigonometry to find the length of \(OP\).

Step 3: Detailed Explanation:
1. In \(\Delta OTP\):
- \(\angle OTP = 90^\circ\) (since radius \(OT \perp\) tangent \(PT\) at the point of contact \(T\))
- \(OT = r\) (radius of the circle)
- \(\angle POT = 45^\circ\) (given)
2. We need to find the hypotenuse \(OP\).
3. Using the cosine trigonometric ratio for \(\angle POT\) in \(\Delta OTP\):
\[ \cos(\angle POT) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OT}{OP} \] 4. Substitute the known values:
\[ \cos 45^\circ = \frac{r}{OP} \] 5. Since \(\cos 45^\circ = \frac{1}{\sqrt{2}}\):
\[ \frac{1}{\sqrt{2}} = \frac{r}{OP} \] 6. Cross-multiplying yields:
\[ OP = r\sqrt{2} \] 7. Therefore, the length of \(OP\) is \(r\sqrt{2}\).

Step 4: Final Answer:
The correct option is (A).