Circle And Tangent

(i) The given statement is false. A tangent is a line which intersect the circle at exactly one point and the point of contact of tangent and circle is called as point of tangency.

The correct statement is as follows:

A secant of a circle is that line which intersects the circle at two points.

(ii) The given statement is false. Since O is the centre of the circle and hence OP is perpendicular to the tangent and the perpendicular is the smallest of all distances.

Therefore OP ≠ OQ.

A tangent XY touches a circle at a point P and Q is any other point on the tangent. If O is the centre of the circle then, OP < OQ.

(iii) The given statement is true. Since the tangent is perpendicular to the diameter, the two tangents LM and XY has to be parallel.

(iv) The given statement is true. Since AOB is a diameter, a semicircle is formed and the angle in a semicircle is always 90° or right angle.

(i) Through a point on a circle one tangent lines can be drawn.

Let the centre be O.

From this figure it can be interpreted that OP will be the radius of the circle.

For the line LM which contains the point P, the distance OP<OL because OP is the perpendicular distance and perpendicular distance is always the shortest distance.

OP is perpendicular to LP and MP.

This is possible only when all the L, P and M lie on same line.

Hence LPM is a straight line.

LM is the only tangent which pass through the point P.

Hence only one tangent can be drawn from a point on a circle.

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two Parallel tangents at the most.

A tangent line touches a shape at a single point. Two parallel tangents will be on opposite sides of the circle. If you try to add another parallel line, it will pass through two points of the circle, and isn't a tangent.

(iv) The common point of a tangent to a circle and the circle is called point of contact or point of tangency.

In the above figure P is the point of contact for the tangent LM and Q is the point of contact for the tangent XY.

Given:

Radius of larger circle = 5 cm.

Radius of smaller circle = 3 cm.

Solution:

The diagram for the question is as follows:

AB is a tangent for smaller circle, and we know that tangent at any point is perpendicular to the radius through point of contact.

Hence, OP ⊥ AB

Also, AB is a chord for large circle, and we know perpendicular from center to the chord bisects the chord

Hence,

Now, In ΔOAP

By Pythagoras theorem,

OA² = AP² + OP²

AP² = OA² – OP²

Here, OA = radius of bigger circle = 5 cm

and OB = radius of smaller circle = 3 cm

AP² = 5² – 3²

AP² = 25 – 9

AP² = 16

∴ AP = 4 cm

Hence,

AB = 2 AP

= 2(4) = 8 cm

Given:

Distance of point from center, OA = 10 cm

Length of tangent, AB = 4 cm.

Solution:

The diagram for the question is as follows:

Let the centre be O.

From this figure it can be interpreted that OB will be the radius of the circle.

Radius OB is perpendicular to the tangent AB as radius is perpendicular to the tangent.

OB ⊥ AB

Hence Δ OAB is right angled triangle.

∠ OBA = 90°

By Pythagoras theorem,

OA² = AB² + OB²

OB² = OA² – AB²

OB² = 10² – 4²

OB² = 100 – 16

OB² = 84

OB² = 21 × 2 × 2

OB = 2√21 cm

Therefore the radius of circle OB = 2√21 cm

Let the diagram be as follows:

P divides the line AB in the ratio 3:1

Adding 1 both side, we get

⇒ BP = 2 cm

And hence,

AP = 3BP = 3(2) = 6 cm

Let the centre be O.

From this figure it can be interpreted that OP will be the radius of the circle.

Radius OP is perpendicular to the tangent AB as radius is perpendicular to the tangent.

If OP is ⊥ to AB which implies that OP is ⊥ AP and OP is ⊥ PB.

Hence Δ OPA is right angled triangle

∠ OPA = 90°

OA² = AP² + OP²

OP² = OA² – AP²

OP² = 10² – 6²

OP² = 100 – 36

OP² = 64

∴ OP = 8 cm

Therefore the radius of circle = 8 cm.

The diagram is as follows:

Let PQRS be the quadrilateral which has a circle inside it with center O.

Now join AO, BO, CO, DO

From the figure, ∠ 1 = ∠ 8 [Two tangents drawn from an external point to a circle subtend equal angles at the centre.]

We can also state that ∠ 2 = ∠ 3, ∠ 4 = ∠ 5 and ∠ 6 = ∠ 7

Sum of angles in a quadrilateral is 360°

Sum of angles at the center is 360°

∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360°

2(1 + 2 + 5 + 6) = 360°

1 + 2 + 5 + 6 = 180°

or

2(8 + 3 + 4 + 7) = 360°

8 + 3 + 4 + 7 = 180°

∠ POQ + ∠ ROS= 180°

Also ∠ QOR + ∠ POS = 180°

[Since ∠ POQ = ∠ 1 + ∠ 2

∠ ROS = ∠ 5 + ∠ 6

∠ QOR = ∠ 7 + ∠ 8

∠ POS = ∠ 3 + ∠ 4

as shown in the figure]

Hence it is proved that PQ and RS subtend supplementary angles at center.

Thus the angles subtended at the centre by a pair of opposite sides are supplementary.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∠ OAP = ∠ OBP = 90°

Now, ∠ OAP + ∠ APB + ∠ OBP + ∠ AOB = 360° [Angle sum property of quadrilaterals]

⟹ 90° + ∠ APB + 90° + ∠ AOB = 360°

⟹ ∠ AOB = 360° - 180° - ∠ APB = 180° - ∠ APB.... (1)

Now, in Δ OAB, OA is equal to OB as both are radii.

⟹ ∠ OAB = ∠ OBA [In a triangle, angles opposite to equal sides are equal]

Now, on applying angle sum property of triangles in Δ AOB,

We obtain ∠ OAB + ∠ OBA + ∠ AOB = 180°

⟹ 2∠ OAB + ∠ AOB = 180°

⟹ 2 ∠ OAB + (180° - ∠ APB) = 180° [Using (1)]

⟹ 2 ∠ OAB = ∠ APB

Thus, the given result is proved

Since tangents to a circle is perpendicular to the radius.

∴ OA ⊥ AP and OB ⊥ BP.

⟹ ∠OAP = 90° and ∠OBP = 90°

⟹ ∠OAP + ∠OBP = 90° + 90° = 180°........... (1)

In quadrilateral OAPB,

∠OAP + ∠APB + ∠AOB + ∠OBP = 360°

⟹ (∠APB + ∠AOB) + (∠OAP + ∠OBP) = 360°

⟹ ∠APB + ∠AOB + 180° = 360° [From (1)]

⟹ ∠APB + ∠AOB = 180°................ (2)

Thus from equations (1) and (2) it can be concluded that the quadrilateral PAOB is a cyclic quadrilateral.

(i) To understand alternate segment following diagram is beneficial:

Angles with same color are equal.

Theorem:

The angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment.

The alternate segment for ∠ BAQ is ∠ ADB.

(ii) The alternate segment for ∠ DAP is ∠ ABD

(iii) The figure is as follows:

When C will be jointed to B then ∠ ACB will be equal to ∠ BAP.

(iv) ∠ABD and ∠ADB are equal to ∠ DAP and ∠ BAQ because angles in alternate segments are equal.

AC is the diameter

∠ ACP = 90° [radius is perpendicular to tangent]

∠ ABC = 90° [angle in a semicircle is 90°]

∠ ABC + ∠ BAC + ∠ ACB = 180° [sum of internal angles in a triangle is 180°]

∠ ACB + 90° + 80° = 180°

∠ ACB = 180 – 170

∠ ACB = 10°

∠ BCP = ∠ ACP - ∠ ACB

= 90 – 10

= 80°

∴ ∠ BCP = 80°

In the given figure,

Let O be the centre of the circle

Join OS

Let O be the centre of the circle

So ∠ SRQ = 90° [radius is perpendicular to tangent]

So ∠ SRT = 90 – 30

= 60°

∠ STR = 90° [angle in a semicircle is 90°]

∠ RSY = 90° [radius is perpendicular to tangent]

∠ RST + ∠ STR + ∠ SRT = 180° [sum of internal angles in a triangle is 180°]

∠ RST + 90 + 60 = 180

∠ RST = 180 – 150

∠ RST = 30°

∠ TSY = ∠ RSY - ∠ RST

= 90 – 30

= 60°

∴ ∠ TSY = 60°

∠ ACD = ∠ ACB …………. (1) [In the question it is given that AC bisects the angle C which means that the two angles ∠ 1 and ∠2 has to be equal]

Here we can see that ∠ PAD = ∠ ABD [Angle in the alternate segment]

Similarly, ∠ QAB = ∠ ADB

Also AB is common arc and ADB and ACB are the angle in the same segment, so

∠ ABD = ∠ ACD

But by the equation (1) we can say that

∠ 1 = ∠ 2

So ∠ PAD = ∠ ADB [Alternate Interior angle]

So PQ ∥ BD