If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
Show Hint
Whenever tangents from an external point are perpendicular to each other, the quadrilateral formed by the center, the two points of contact, and the external point is a square. The distance from the center to the external point is always \(r\sqrt{2}\).
Step 1: Understanding the Concept:
A tangent to a circle is perpendicular to the radius through the point of contact.
The length of tangents drawn from an external point to a circle are equal (\(PQ = PR\)).
In a quadrilateral, if three angles are \(90^{\circ}\) and adjacent sides are equal, it forms a square. Step 2: Key Formula or Approach:
In quadrilateral \(OQPR\):
\[ \angle OQP = \angle ORP = 90^{\circ} \text{ (Radius } \perp \text{ Tangent)} \]
\[ \angle QPR = 90^{\circ} \text{ (Given)} \]
Since the sum of angles is \(360^{\circ}\), \(\angle QOR = 90^{\circ}\).
Also, \(OQ = OR = 4 \text{ cm}\) (Radii).
Thus, \(OQPR\) is a square with side length \(s = 4 \text{ cm}\). Step 3: Detailed Explanation:
In the square \(OQPR\), \(OP\) is the diagonal.
The length of the diagonal of a square with side \(s\) is given by \(s\sqrt{2}\).
\[ OP = 4\sqrt{2} \text{ cm} \]
Alternatively, using Pythagoras theorem in \(\triangle OQP\):
\[ OP^2 = OQ^2 + QP^2 \]
Since \(OQPR\) is a square, \(QP = OQ = 4 \text{ cm}\).
\[ OP^2 = 4^2 + 4^2 = 16 + 16 = 32 \]
\[ OP = \sqrt{32} = 4\sqrt{2} \text{ cm} \] Step 4: Final Answer:
The length of \(OP\) is \(4\sqrt{2} \text{ cm}\).
