If the angle between two tangents drawn from a point \(P\) to a circle of radius \(a\) and centre \(O\) is \(90^\circ\), then \(OP = a\sqrt{2}\). State True/False and justify.
True.
Step 1: Draw a circle with centre \(O\) and radius \(a\). Take a point \(P\) outside the circle. Draw two tangents from \(P\) that touch the circle at points \(A\) and \(B\).
Step 2: The angle between the tangents at point \(P\) is given as \(90^\circ\). This means \(\angle APB = 90^\circ\).
Step 3: Join \(O\) to \(A\), \(O\) to \(B\), and \(O\) to \(P\). We now have quadrilateral \(OAPB\).
Step 4: The angle at the centre is related to the angle between the tangents. Formula: \(\angle AOB = 180^\circ - \angle APB\). Substituting \(\angle APB = 90^\circ\): \[ \angle AOB = 180^\circ - 90^\circ = 90^\circ. \]
Step 5: Triangle \(AOB\) is isosceles because \(OA = OB = a\) (both are radii). So, \(\angle AOB = 90^\circ\). Each of the other two angles at \(A\) and \(B\) = \(45^\circ\).
Step 6: Consider right triangle \(AOP\). Here, \(OA = a\) (radius), \(\angle AOP = 45^\circ\).
Step 7: Apply the definition of cosine: \[ \cos(\angle AOP) = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{OA}{OP} \] Substituting values: \[ \cos 45^\circ = \dfrac{a}{OP}. \]
Step 8: We know \(\cos 45^\circ = \tfrac{1}{\sqrt{2}}\). So, \[ \dfrac{1}{\sqrt{2}} = \dfrac{a}{OP}. \]
Step 9: Cross multiply: \[ OP = a \times \sqrt{2}. \]
Final Result: The distance from the centre to the external point is \[ OP = a\sqrt{2}. \] Hence, the statement is True.