Theorem-1 :- The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution :-
Given that : A circle with centre O and a tangent XY to the circle at a point P.
Prove that : OP is perpendicular to XY.
Construction : Take a point Q on XY other than P and join OQ.
Proof : The point Q must lie outside the circle.
Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle).
Therefore, OQ is longer than the radius OP of the circle. i.e., OQ > OP.
Since this happens for every point on the line XY except the point P,
OP is the shortest of all the distances of the point O to the points of XY.
So OP is perpendicular to XY.
Theorem-2 :- The lengths of tangents drawn from an external point to a circle are equal.
Solution :-
Given that : A circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P.
Prove that : PQ = PR.
Construction : we join OP, OQ and OR.
Proof : ∠ OQP and ∠ ORP are right angles, because these are angles between the radii
and tangents, and according to Theorem 10.1 they are right angles.
Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore, Δ OQP ≅ Δ ORP (RHS)
This gives PQ = PR (CPCT).
