The length of the tangent from an external point \(P\) on a circle with centre \(O\) is always less than \(OP\). State True/False and justify.
True.
Step 1: Let the circle have centre \(O\) and radius \(r\). Take an external point \(P\) outside the circle.
Step 2: Draw a tangent from \(P\) to the circle, touching the circle at point \(T\).
Step 3: In geometry, a radius drawn to the point of tangency is always perpendicular to the tangent. So, \(OT \perp PT\).
Step 4: This makes triangle \(OPT\) a right-angled triangle at \(T\).
Step 5: By Pythagoras theorem: \[ OP^{2} = OT^{2} + PT^{2} \]
Step 6: Substituting, \(OT = r\) (radius), we get: \[ OP^{2} = r^{2} + PT^{2} \]
Step 7: Rearranging: \[ PT^{2} = OP^{2} - r^{2} \]
Step 8: Taking square root: \[ PT = \sqrt{OP^{2} - r^{2}} \]
Step 9: Since \(r^{2} > 0\), we see that: \(PT < OP\).
Final Result: The tangent length from \(P\) (i.e., \(PT\)) is always smaller than the distance \(OP\). Hence the statement is True.