If a circle and a line are drawn on a paper, three things can possible happen:

t1(i) The line does not touches or cuts the circle. As you see in the figure besides that XY does not touch or cut the circle.

(ii) The line cuts the circle in two parts. The straight line which cuts the circle in two points is called the secant of the circle. In this case the line PS cuts the circle at two points Q and R . QR is a chord.

(iii) The line touches the circle at only one point. The line that touches the circle at only one point is called tangent of the circle. The point at which the line touches the circle is called the point of contact. In this case, you see that the line AB touches the circle at point C (one point only).

Theorems related to tangents:

t3Theorem 18: The tangent at any point of a circle and the radius through this point are perpendicular to each other.

Given: AB is a tangent. Point of contact is P .

To prove: OP \perp AB

Proof: OP <  OQ (since Q is outside the circle)

Similarly, we can show that out of all possible line segments that could be drawn from O to the line AB, OP would be the shortest. Hence OP \perp AB .  (Reason: The shortest line segment, drawn from a given point to a given line is perpendicular to this line)

Hence Proved

t4Corollary: If two tangents are drawn to a circle from an external point, then

(i) the lengths of the tangents to the circle are equal

(ii) the tangents will subtend equal angles at the center of the circle

(iii) tangents are equally inclined to the line joining the point and the center of the circle.

Given: A circle with center O . PA and PB are two tangents to this circle from external point P .

To Prove:

(i) PA = PB

(ii) \angle AOP = \angle BOP

(iii) \angle APO = \angle BPO

Proof:

Consider \triangle AOP and \triangle BOP

OA = OB (radius of the same circle)

\angle OAP = \angle OBP = 90^o (Theorem 18)

OP is common

\therefore \triangle AOP \cong \triangle BOP (by R.H.S postulate)

Since corresponding parts of the congruent triangles are equal, we get

(i) PA = PB

(ii) \angle AOP = \angle BOP

(iii) \angle APO = \angle BPO

Hence Proved

Theorem 19: If two circles touch each other, the point of contact lines on the straight line through the centers.

There are two possible scenarios

t5Case 1: when the circles just touch each other externally

Given: Two circles with centers A and B touch each other externally at point P as shown in the diagram.

To Prove: P lies on the line AB i.e. A, P and B are collinear.

Proof:

\angle APQ = 90^o (angle between the radius and the tangent)

\angle BPQ = 90^o (angle between the radius and the tangent)

\angle APQ \angle BPQ = 180^o

\Rightarrow \angle APB = 180^o which means that APB is a straight line.

Hence Proved

t6Case 2: when the two circles touch each other internally

Given: Two circles with centers A and B touch each other internally at point P as shown in the diagram.

To Prove: P lies on the line AB i.e. A, B and P are collinear.

Proof:

\angle APQ = 90^o (angle between the radius and the tangent)

\angle BPQ = 90^o (angle between the radius and the tangent)

Therefore both AP and BP are perpendicular to the tangent PQ at the point P .

Therefore AP and BP lie on the same line because only one perpendicular can be drawn through a line through a point on it.

Hence Proved

Theorems related to Chords:

Theorem 20: If two chords of a circle intersect internally or externally then the product of the lengths of their segments is equal.

There are two possible cases.

t10Case 1: When the chords intersect internally

Given: Chords AB and CD of a circle intersect each other at point P inside the circle.

To Prove: PA \times PB = PC \times PD

Proof: Consider \triangle APC and \triangle PBD

\angle CAB = \angle CDB (angles in the same segment)

\angle ACD = \angle DBA (angles in the same segment)

Therefore \triangle APC \sim \triangle PBD (AAA Postulate)

\Rightarrow \frac{PA}{PD} =\frac{PC}{PB} (corresponding sides of similar triangles are proportional)

\Rightarrow PA \times PB = PC \times PD

Hence Proved

t7Case 2: When the chords intersects externally

Given: Chords AB and CD of a circle, when produced,  intersect each other at point P outside the circle.

To Prove: PA \times PB = PC \times PD

Proof: Consider \triangle PAC and \triangle PDB

\angle PAC = \angle PDB (external angle of a cyclic quadrilateral is equal to internal opposite angle)

\angle ACP = \angle PBD  (external angle of a cyclic quadrilateral is equal to internal opposite angle)

\triangle PAC \sim \triangle PDB (By AAA postulate)

\Rightarrow \frac{PA}{PD} =\frac{PC}{PB} (corresponding sides of similar triangles are proportional)

\Rightarrow PA \times PB = PC \times PD

Hence Proved

t8Theorem 21: The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

Given: A circle with center O . Tangent PQ touches the circle at point A . AB is a chord drawn through point A .

To Prove: \angle BAQ = \angle ACB  and \angle BAP = \angle ADB

Proof: In \triangle ABR

\angle ABR = 90^o  (angle in a semi circle)

\Rightarrow \angle ARB + \angle RAB = 90^o

\angle OAQ = 90^o (angle between the radius and the tangent)

\Rightarrow \angle RAB + \angle BAQ = 90^o

\angle ARB + \angle RAB = \angle RAB + \angle BAQ

\Rightarrow \angle ARB = \angle BAQ

But \angle ARB = \angle ACB (angles in the same segment)

\therefore \angle BAQ = \angle ACB

Now \angle BAP + \angle BAQ = 180^o (PAQ is a straight line)

\angle ACB + \angle ADB = 180^o (Opposite angles of a cyclic quadrilateral)

\Rightarrow \angle BAP + \angle BAQ = \angle ACB + \angle ADB

\Rightarrow \angle BAP = \angle ADB as \angle BAQ = \angle ACB

Hence Proved

t9.jpgTheorem 22: If a chord and a tangent intersect externally, then the product of the length of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

Given: Chord AB and tangent TP of a circle intersect each other at point P outside the circle.

To Prove: PA \times PB = PT^2

Proof: Consider \triangle PAT and \triangle PTB

\angle PTB = \angle TAB (Angles in alternate segment)

\angle TPA = \angle TPB (common angle)

\triangle PAT \sim \triangle PTB (AAA Postulate)

\Rightarrow \frac{PA}{PT} =\frac{PT}{PB} (corresponding sides of similar triangles are proportional)

\Rightarrow PA \times PB = PT^2

Hence Proved

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