Question

If PQ and PR are tangents to the circle with centre O and radius 4 cm such that \(\angle QPR = 90^\circ\), then the length OP is

• A. 4 cm
• B. \(4\sqrt{2}\) cm
• C. 8 cm
• D. \(2\sqrt{2}\) cm

Hint:

Whenever the angle between two tangents drawn from an external point to a circle of radius \(r\) is \(90^\circ\), the quadrilateral formed by the radii and the tangents is always a square.
The distance of the external point from the center of the circle is simply the diagonal of this square, which is always \(r\sqrt{2}\).
This allows you to write down the answer directly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The topic of this question is Circles and Tangents.
A key theorem in circle geometry states that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
We are given a circle centered at \(O\) with a radius of \(4\) cm.
Two tangents, \(PQ\) and \(PR\), are drawn from an external point \(P\) to touch the circle at points \(Q\) and \(R\) respectively.
The angle between the two tangents is given as \(\angle QPR = 90^\circ\).
We need to determine the length of the line segment \(OP\) connecting the center of the circle to the external point \(P\).

Step 2: Key Formula or Approach:
Since \(PQ\) and \(PR\) are tangents, the radii \(OQ\) and \(OR\) are perpendicular to the tangents at the points of contact:
\[ \angle OQP = 90^\circ \] \[ \angle ORP = 90^\circ \] We can analyze the geometry of the quadrilateral \(OQPR\) and use the properties of right-angled triangles and the Pythagoras theorem to find the length of \(OP\).

Step 3: Detailed Explanation:

• Let us look at the quadrilateral \(OQPR\):
The sum of the interior angles of a quadrilateral is \(360^\circ\).
We know that:
\(\angle OQP = 90^\circ\) (angle between radius \(OQ\) and tangent \(PQ\))
\(\angle ORP = 90^\circ\) (angle between radius \(OR\) and tangent \(PR\))
\(\angle QPR = 90^\circ\) (given)

• Calculate the remaining angle \(\angle QOR\):
\[ \angle QOR = 360^\circ - (\angle OQP + \angle ORP + \angle QPR) \] \[ \angle QOR = 360^\circ - (90^\circ + 90^\circ + 90^\circ) \] \[ \angle QOR = 360^\circ - 270^\circ = 90^\circ \]

• Identify the shape of quadrilateral \(OQPR\):
Since all four angles are \(90^\circ\), \(OQPR\) is a rectangle.
Additionally, the adjacent sides \(OQ\) and \(OR\) are equal because they are both radii of the same circle (\(OQ = OR = 4\) cm).
A rectangle with equal adjacent sides is a square.
Therefore, the quadrilateral \(OQPR\) is a square of side \(4\) cm.
This means the lengths of the tangents are also equal to the radius:
\[ PQ = PR = 4 \text{ cm} \]

• Consider the right-angled triangle \(\Delta OQP\):
Apply the Pythagoras theorem to \(\Delta OQP\):
\[ OP^2 = OQ^2 + PQ^2 \] Substitute the values \(OQ = 4\) cm and \(PQ = 4\) cm:
\[ OP^2 = 4^2 + 4^2 \] \[ OP^2 = 16 + 16 = 32 \] \[ OP = \sqrt{32} \] \[ OP = \sqrt{16 \times 2} = 4\sqrt{2} \text{ cm} \]

Step 4: Final Answer:
The length of the segment \(OP\) is \(4\sqrt{2}\) cm.
Therefore, the correct option is (B).