Chapter Notes: Circle CBSE Class 10



Notes for circles chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

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(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: OP \bot AB
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 OP \bot AB, 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
 \Rightarrow        OR" />
 \Rightarrow        OP" />        [OP=OR]
 \Rightarrow       OQ < OQ
Thus, OP is shorter than any other segment joining O to any point of AB.
Hence, OP \bot AB.

(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 \Delta OPA \cong \Delta OQA.
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OP \bot AB and OQ \bot AQ
 \Rightarrow \angle OPA = \angle OQA = {90^ \circ }....(i)
Now, in right triangle OPA and OQA, we have
OP=OQ                        [Radii of a circle]
\angle OPA = \angle OQA             [From (i)]
And,             OA=OA                        [Common]
So, by RHS-Criterion of congruence, we get
\Delta OPA \cong \Delta OQA
 \Rightarrow         AP = AQ

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