# 10th Class Mathematics Circles Introduction

Introduction

Category : 10th Class

### Introduction

Circle is defined as the locus of a point which is at a constant distance from a fixed point. The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.

### Tangent to a circle

A tangent to a circle is a line which intersects the circle at exactly one point. The point where the tangent intersects the circle is known as the point of contact.

Properties of tangent to a circle

Following are some properties of tangent to a circle:

• A tangent to a circle is perpendicular to the radius through the point of contact.
• A line drown through the end-point of a radius and perpendicular to it is a tangent to the circle.
• The length of he two tangents drawn from an external point to a circle are equal.
• If two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre.
• If two tangents are drawn to a circle from an external point, then they are equally inclined to the segment, joining the centre to that point.

From the above points we conclude that in the following figure;

$\angle OPT=\angle OQT={{90}^{o}},\,\angle POT=\angle QOT$ $\angle QTO=\angle OTP\,and\,PT\,=\,QT$

Two tangents PT and QT are drawn to a circle with centre 0 from an external point as shown in the following figure, then:

(a) $\angle QTP=\angle QPO$

(b) $\angle QTP=2\angle QPO$

(c) $\angle QTP=3\angle QPO$

(d) $\angle QTP={{90}^{o}}$

(e) None of these

Explanation

In the given figure, we have

TP = TO.            [tangents drawn from an external point are equal in length]

$\Rightarrow \,\,\angle TPQ=\,\angle TPQ$

$In\,\angle QTQ,\,we\,\,have$

$\angle TPQ+\angle TQP+PTQ={{180}^{o}}$

$\Rightarrow \,\angle TPQ={{90}^{o}}-\frac{1}{2}\angle PTQ$

$\Rightarrow \,\frac{1}{2}\angle PTQ={{90}^{o}}-\angle TPQ$     .... (i)

Also, $\angle OPT={{90}^{o}}$

$\Rightarrow \,\angle OPQ={{90}^{o}}-\angle TPQ$            .... (ii)

From (i) and (ii), we get

$\frac{1}{2}\angle PTQ=\angle OPQ\Rightarrow \angle PTQ=2\angle OPQ$

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