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).