**Circle:** A circle is a collection of all points in a plane which are at a constant distance from a fixed point.

**Centre:** The fixed point is called the centre.

**Radius:** The constant distance from the centre is called the radius.

**Chord:** A line segment joining any two points on a circle is called a chord.

**Diameter:** A chord passing through the centre of the circle is called diameter. It is the longest chord.

**Tangent:** When a line meets the circle at one point or two coincidings The line is known as points, a tangent.

The tangent to a circle is perpendicular to the radius through the point of contact.

⇒ OP ⊥ AB

## Theorems

### Tangent perpendicular to the radius at the point of contact

**Theorem 1**: The theorem states that “the **tangent **to the circle at any point is the **perpendicular to the radius** of the circle that passes through the point of contact”.

**Given:** XY is a tangent at point P to the circle with centre O.

**To prove:** OP ⊥ XY

**Construction:** Take a point Q on XY other than P and join OQ

**Proof:** If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle

OQ > OP

This happens with 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

OP ⊥ XY …[Shortest side is the perpendicular]

**Theorem 2:** A line drawn through the end point of a radius and perpendicular to it, is the tangent to the circle.

**Given:** A circle C(O, r) and a line APB is perpendicular to OP, where OP is the radius.

**To prove:** AB is tangent at P.

**Construction:** Take a point Q on the line AB, different from P and join OQ.

**Proof:** Since OP ⊥ AB

OP < OQ ⇒ OQ > OP

The point Q lies outside the circle.

Therefore, every point on AB, other than P, lies outside the circle.

This shows that AB meets the circle at point P.

Hence, AP is a tangent to the circle at P.

**Theorem 3:**Prove that the lengths of tangents drawn from an external point to a circle are equal

**Given:**PT and PS are tangents from an external point P to the circle with centre O.

**To prove:**PT = PS

**Construction:**Join O to P, T and S.

**Proof:**In ∆OTP and ∆OSP.

OT = OS …[radii of the same circle]

OP = OP …[common]

∠OTP = ∠OSP …[each 90°]

∆OTP = ∆OSP …[R.H.S.]

PT = PS …[c.p.c.t.]

If two tangents are drawn to a circle from an external point, then:

- They subtend equal angles at the centre i.e., ∠1 = ∠2.
- They are equally inclined to the segment joining the centre to that point i.e., ∠3 = ∠4.

∠OAP = ∠OAQ

NCERT Solutions Ch- 10 Maths