Eduraworld : A learning guide

Circles

(1) Prove that Tangent to a circle at a point is perpendicular to the radius through the point of contact.Given: A circle C(O,r) and a tangent AB at a point P.
To Prove: OPAB
Construction: Take any Point Q, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.Proof: We know that among all line segments joining the point O to a point on AB the shortest one is perpendicular to AB. So, to prove that OPAB, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Cleraly,      OP=OR               [Radii of the same circle]
Now,          OQ=OR+RQ
       OQ>OR
       OQ>OP        [OP=OR]
       OQ<OQ
Thus, OP is shorter than any other segment joining O to any point of AB.
Hence, OPAB.
(2) Prove that from a point, lying outside a circle, two and only two tangents can be drawn to it.When the point lies outsides the circle, there are exactly two tangents to circle from a point which lies outside the circle. As shown in figure.
(3) Prove that the lengths of the two tangents drawn from an external point to a circle are equal.Given: AP and AQ are two tangents from a point A to a circle C(O,r).
To Prove: AP=AQ
Construction: Join OP,OQ and OA
Proof: In order to prove that AP=AQ we shall first prove that ΔOPAΔOQA.
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OPAB and OQAQ
OPA=OQA=90....(i)
Now, in right triangle OPA and OQA, we have
OP=OQ                        [Radii of a circle]
OPA=OQA             [From (i)]
And,             OA=OA                        [Common]
So, by RHS-Criterion of congruence, we get
ΔOPAΔOQA
         AP=AQ

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