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 from the center to the external point is simply the diagonal of this square, which is calculated as \(r\sqrt{2}\).
Recognizing this immediately allows you to bypass the calculation!
Step 1: Set up the right triangle.
Since $PQ$ is a tangent at $Q$, the radius $OQ$ is perpendicular to $PQ$, so triangle $OQP$ is right angled at $Q$, with $OQ = 4$ cm. Step 2: Use the angle bisector property.
$OP$ bisects $\angle QPR$ because $PQ$ and $PR$ are tangents of equal length from $P$, so $\angle QPO = \frac{90^\circ}{2} = 45^\circ$. Step 3: Apply a trigonometric ratio instead of Pythagoras.
In right triangle $OQP$, $\sin(\angle QPO) = \frac{OQ}{OP}$, so $\sin 45^\circ = \frac{4}{OP}$.
Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we get $OP = 4\sqrt{2}$ cm.
\[ \boxed{OP = 4\sqrt{2}\ \text{cm}} \]
