# 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.

(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,          $OQ=OR+RQ$
$\Rightarrow$       $OQ > OR$
$\Rightarrow$       $OQ > OP$        [$OP=OR$]
$\Rightarrow$       $OQ < OQ$
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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