Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.


Analysis: As shown in the above figure, let line l be the tangent to the circle at point A. Line AB is a chord and
BCA is an inscribed angle. Now by tangent- secant angle theorem, BCA BAR.
By converse of tangent- secant theorem, if we draw the line l such that, BAR BCA, then it will be the required tangent.
Rough figure is shown below.

Construction steps:
(1) Draw a circle of any radius . Take any point A on it.
(2) Draw chord AB and an inscribed BCA .
(3) With the centre C and any convenient radius draw an arc intersecting the sides of BCA in points P and Q.
(4) Using the same radius and centre A, draw an arc intersecting the chord AB at point S.
(5) Taking the radius equal to d(PQ) and centre S, draw an arc intersecting the arc drawn in the previous step. Let R be the point of intersection of these arcs. Draw line AR. Line AR is the required tangent to the circle.

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