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
Show 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!
Step 1: Use symmetry of the two tangents. Since PQ and PR are tangents from the same external point P, PO bisects the angle \(\angle QPR\), so \(\angle OPQ = \angle OPR = 45^\circ\). Step 2: Use the tangent-radius right angle. The radius \(OQ\) is perpendicular to tangent \(PQ\), so triangle \(OQP\) is right angled at Q, with \(\angle OPQ = 45^\circ\) and \(OQ = 4\) cm. Step 3: Apply a trig ratio in this right triangle. Using \(\sin(\angle OPQ) = \frac{OQ}{OP}\): \[ \sin 45^\circ = \frac{4}{OP} \] \[ \frac{1}{\sqrt{2}} = \frac{4}{OP} \implies OP = 4\sqrt{2} \text{ cm} \] Step 4: Confirm the final answer. The length OP is \(4\sqrt{2}\) cm, matching option (B). \[ \boxed{OP = 4\sqrt{2} \text{ cm}} \]
