## Circles

In this third part of the Geometry basic and rich concepts tutorial series we will cover Circles. If you like, you may refresh your knowledge on points, lines and triangles by referring to the * first part* and on quadrilateral, rectangles and squares by referring to the

*of the geometry tutorial series.*

**second part**A circle is a closed curve in which every point on the curve is equidistant from a single point inside the enclosed area. The central point is the **center**, the distance from the center to any point on the closed curve is the **radius** and the closed curve is a **circle**.

A unique circle can be defined by two parameters: namely, the location of its center and the length of its radius $r$.

A line connecting two points on the circle and passing through the center is of length double the radius and is the **diameter** $d$ of the circle.

Diameter $d=2r$.

### Chord of a circle

Chord is a line connecting two points on a circle. At the minimum it can be a **point** with two points coinciding and at the maximum it can be the **diameter** passing through the center.

In the figure below $AB$ is a general chord.

#### Properties of a chord:

- The perpendicular from the center of a circle $O$ to a chord $AB$ always
**bisects the chord $AB$**and**the angle included by the chord at the center $\angle AOB$**.

**Reason:** Being the radii, $AO=BO$ and perpendicular side $OP$ is common between the two right triangles $\triangle AOP$ and $\triangle BOP$. Thus the third side is also equal (by Pythagoras theorem) and these two triangles are congruent. So, corresponding sides $AP=PB$ and corresponding angles $\angle AOP=\angle BOP$.

- Chords of equal length $AB$ and $CD$ subtend the same angle at the center.

If the chord $AB$ is rotated about the center, the triangle $\triangle AOB$ rotates. Eventually, as $AB=CD$, $\triangle AOB$ can be perfectly superimposed on the $\triangle COD$.

**Second reasoning:** In triangles $\triangle AOB$ and $\triangle COD$, three sides are equal and thus these are congruent. So angles subtended by the two chords at the center are equal.

- The triangle $\triangle AOB$ subtended by the chord $AB$ at the center is always an isosceles triangle.

- As chords $AB=CD$ the arcs $AB$ and $CD$ are also equal. Equal chords are made by equal arcs in a circle.

- If a pair of chords intersect (inside or outside of a circle) the product of the intercepts are equal, that is, $AE\times{CE}=BE\times{DE}$. We classify this as a
**rich concep**t and will shortly see the reason behind it.

### Subtended angles

We will compare here the angles subtended by chords at the center and on a point on the circle.

- Angle subtended by a semicircle or a diameter is always $90^0$.

Being radii, $OA=OB=OP$ in two triangles $\triangle APO$ and $\triangle BPO$. Angle $\angle POA$ being a right angle, in the two isosceles triangles $\triangle APO$ and $\triangle BPO$, all the angles $\angle OAP=\angle OPA=\angle OPB=\angle OBP=45^0$. This makes $\angle APB=90^0$ and $PO$ an angle bisector.

#### Angle subtended by an arc or a chord:

An arc (or its chord) subtends an angle at the center and another angle at any other point of the remaining arc. The second angle is constant and is half the angle subtended by the arc at the center.

There are two concepts here: first, the arc $AB$ subtends the same angle at any point on the rest of the periphery of the circle; and second, the angle subtended by the arch $AB$ at the center is twice this angle subtended at the periphery.

For proof of this * arc angle subtending concept,* you may refer to the discussion,

**Basic and rich Geometry concepts part 4, proof of arc angle subtending concept.**### Rich concept of Cyclic Quadrilateral

We have already mentioned in our previous session on geometry that if four vertices of a quadrilateral lie on the circumference of a circle, the sum of its opposite angles is always $180^0$ and the quadrilateral is called a cyclic quadrilateral. We will see now how it happens.

In the process we will also learn about another rich concept of angle subtended by complementary or alternate arc segment. Explanations will be based on the following figure.

Let us focus our attention on the chord $BD$ and the angles subtended by it.

- It subtends an angle of $\theta$ at $C$ on the circumference and opposite side of the center.
- It subtends and angle of $2\theta$ at the center $O$ which is double the angle subtended at $C$.

We know these two concepts.

Now the **first rich concept** we will encounter is the **concept of angle subtended by the complementary arc**. Angle subtended by a chord is no longer sufficient to express the new situation satisfactorily.

**Angle subtended by complementary arc**

The previous two subtended angles that we listed were subtended by the chord $BD$ and also by the small arc $BAD$. Its complementary arc is the large rest portion of the circumference of the circle, that is, the arc $BCD$. The chord is same but the arcs are different. It is natural as we know,

A chord is jointly held by two arcs of a circle that together form the whole of the circumference.

and that's why we call them **complementary arcs**.

The angle subtended by the complementary arc $BCD$ at point $A$ is $\alpha$ and corresponding angle held at the center is $2\alpha$. So we have the rest two subtended angles held by the complementary arc as,

- The angle $\alpha$ held at $A$ and,
- The angle $2\alpha$ subtended by the same supplementary arc at the center.

Now we can easily see that the total of opposite angles of the quadrilateral, $\alpha + \theta$ is half of total of two subtended angles at the center, that is, half of $2\alpha + 2\theta = 360^0$, which is, $180^0$.

### Rich concept of relation between intercepts when two chords intersect

Let us reproduce the figures of intersection for convenience.

#### Chords intersecting outside the circle

In the first figure above the two chords $AC$ and $BD$ intersect at $E$ outside the circle.

In the triangle $\triangle CDE$, the angle $\angle ACD$ is the external angle and so, $\angle ACD = 180^0 - \angle DCE$.

Again from the property of opposite angles of a cyclic quadrilateral we have,

$\angle ACD = 180^0 - \angle ABD$.

Equating the two we have, $\angle DCE = \angle ABD$.

The common vertex angle at $E$ being same in the two triangles, $\triangle DCE$ and $\triangle ABE$, the third angle must also be same and the two triangles are similar.

Thus from the property of equality of ratios of corresponding sides of two similar triangles, we have,

$\displaystyle\frac{AE}{DE} = \displaystyle\frac{BE}{CE}$

Or, $AE\times{CE} = BE \times{DE}$.

#### Chords intersecting inside the circle

In the second figure above, the chords $AC$ and $BD$ intersect inside the circle at $E$.

In the two triangles $\triangle ADE$ and $\triangle BCE$, the angles at $E$ are equal and also angles $\angle A$ and $\angle B$ held by the arc $DC$ are equal making the two triangles similar.

Thus again the ratio of corresponding sides of the two triangles are equal and we have,

$\displaystyle\frac{AE}{BE} = \displaystyle\frac{DE}{CE}$

Or, $AE\times{CE} = BE \times{DE}$.

### Tangents

When a straight line is drawn through a circle, it intersects the circle at two points. But if the line is rotated towards any side gradually, the resulting chord length gets reduced. Ultimately when the line touches at only one point of the circle, the resulting chord length gets zero and in this state we name the line a **tangent** of the circle.

$A$ is an external point to the circle with center at $O$. The straight line $AD$ cuts a chord from the circle intersecting at two points. The line is rotated left to $AB$. It still intersects the circle at two points but the chord cut off from the circle is now reduced in length.

As the line is rotated still further gradually, a position is reached when it touches the circle at just one point $P$.

At this point $P$, the line $AP$ is a **tangent** to the circle.

#### Basic properties of a tangent:

- A basic property of a tangent is: the radius passing through the tangent point is always perpendicular to the tangent line. Thus angle $\angle APO$ is $90^0$.

- There are always two tangents to a circle from an external pointy $A$. $AQ$ is the second tangent here.

- The line $AO$ from the external point $A$ to the center $O$ bisects the angle held by the two tangents - angle $\angle PAQ$, as also bisects the angle held by the perpendiculars at the center - angle $\angle POQ$ and divides the quadrilateral $APOQ$ into two congruent triangles.

- The lengths of the two tangents $AP=AQ$.

**Reason: **

In two right triangles $\triangle APO$ and $\triangle AQO$, the radii $PO=QO$ and side $AO$ is common. So by Pythagoras theorem, the third sides must also be equal, that is, $AP=AQ$. This makes the two triangles congruent and the mid-line $AO$ bisects both the angles $\angle PAQ$ and $\angle POQ$.

**External angle of a tangent meeting a chord equals angle subtended by the chord in the alternate segment of circumference - a rich geometric concept:**

The external angle made by a chord with a tangent at the point of tangency is equal to any angle subtended by the chord at any point in the alternate segment of the circle.

As we already know, the angle subtended by a chord is same for any point on the alternate segment.

Angle $\angle APB$ is the external angle to the chord $BP$ at the tangent point.

$ \angle APB=\angle PCB=\angle PDB=\theta=\frac{1}{2}\angle POB $.

**Reason:**

As $OP$ is a perpendicular on $AP$ at $P$,

$\angle OPB = 90^0 - \theta$.

Being two radii, $OP = OB$ and triangle $\triangle POB$ is isosceles. Thus the base angles in this triangle are equal,

$\angle OBP = \angle OPB = 90^0 - \theta$.

So in triangle $\triangle OPB$,

$\angle POB = 180^0 - (\angle OBP + \angle OPB)$

$\hspace{14mm}=180^0 - (180^0 - 2\theta)$

$\hspace{14mm}=2\theta$.

This angle being the angle held by the chord $PB$ at center $O$, it will be twice of angle subtended by the chord at any point in the alternate segment $PDCB$. The subtended angles will then be equal to $\theta$ and hence will equal the external angle made by the chord $BP$ at tangent point $P$ with the tangent $AP$.

#### Secant and tangent of a circle - equality of product of intercepts, a rich geometric concept:

When the line $ABC$ starting from an external point $A$ cuts the circle at two points $B$ and $C$ it is called a **secant** of the circle.

With the line $AP$ tangent to the circle at point $P$, we have the relation:

$AP^2=AB\times{AC} $

**Reason:**

By the rich geometric concept of square of tangent segment $AP$ equaling product of secant segments $AB$ and $AC$, we have,

$AP^2 = AB\times{AC}$.

We need to prove this relationship.

Whenever we face such product equality, we know immediately that there must be two pairs of sides proportional to each other in two similar triangles.

This is a **problem solving rule** that we invariably apply first in such cases.

Searching for two such prospective triangles that would turn out to be similar, we identify $\triangle APB$ and $\triangle APC$ as the candidate pair of triangles by analyzing the sides involved.

In these two triangles, $\angle APB$ is the **external angle** **to chord $PB$** and so will be **equal to the $\angle PCB$ subtended by the chord on the alternate segment**. The second $\angle A$ is common to the two triangles and so **these two triangles are similar** as expected.

This gives us **ratio equality of corresponding pairs of sides,**

$\displaystyle\frac{AP}{AC} = \displaystyle\frac{AB}{AP}$,

Or, $AP^2 = AB\times{AC}$.

In the equation above, the first $AP$ in the numerator is from the smaller triangle and the second $AP$ in the denominator is from the larger triangle.

**The rule we apply,**

Identify the corresponding angles in a similar triangle to identify the corresponding sides that are opposite to the corresponding angles.

These form **a small a core set of concepts on Circles.** Though the concepts are not many, a large number of problems can be solved by applying these concepts along with the other basic concepts covered in our * Geometry basic concepts part 1* and

**part 2.****Note:** You might have noticed that we identified a few concepts as **rich concepts.** This in fact is the indication that new results may need to be derived from the basic concepts, the way the rich concepts have been derived above from the basic concepts.

We classify all concepts into two layers, the **basic concept layer** and the **rich concept layer**. We go on including the new derived concepts in the rich concept layer as and when we discover such a concept. There can be many rich concepts.

Usually all rich concepts can be derived quickly from basic concepts if you know how. **Nevertheless, applying a rich concept to solve a problem directly increases the speed of problem solving considerably.**

We will now end this session with a set of problems for you to solve as exercise.

### Problem exercise

The recommended time limit is 18 minutes

#### Problem 1.

In a right triangle $\triangle ABC$, the $\angle A = 90^0$ and $AD$ is perpendicular to $BC$. If areas of the triangles $\triangle ABC = 40cm^2$ and $\triangle ACD = 10cm^2$ with $AC = 9cm$, the length of $BC$ is,

- 4cm
- 12cm
- 18cm
- 6cm

#### Problem 2.

If the ratio of an external angle and an internal angle of a regular polygon is $1 : 17$, the number of sides of the regular polygon is,

- 12
- 36
- 20
- 18

#### Problem 3.

In an isosceles $\triangle ABC$, $AB=AC$. A circle passing through $B$ and touching AC at its middle point, intersects $AB$ at $P$. Then $AP : AB$ is,

- 2 : 3
- 4 : 1
- 1 : 4
- 3 : 5

#### Problem 4.

In $\triangle ABC$, $\angle C$ is an obtuse angle. The bisectors of exterior angles at $A$ and $B$ meet extended $BC$ and $AC$ at $D$ and $E$ respectively. If $AB = AD=BE$ then $\angle ACB$ is

- $105^0$
- $110^0$
- $135^0$
- $108^0$

#### Problem 5.

How many triangles can be formed by taking any three from the four line segments of lengths, 2cm, 3cm, 5cm and 6cm?

- 1
- 2
- 3
- 4

#### Problem 6.

In right $\triangle ABC$, $BL$ and $CM$ are two medians with right angle at $\angle A$ and $BC=5cm$. If $BL = \displaystyle\frac{3\sqrt{5}}{2}$, then the length of $CM$ is,

- $2\sqrt{5}$cm
- $10\sqrt{2}$cm
- $5\sqrt{2}$cm
- $4\sqrt{5}$ cm

#### Problem 7.

In $\triangle ABC$, $D$ and $E$ are two points on the sides $AC$ and $BC$ respectively, such that $DE=18$cm, $CE=5$cm and $\angle DEC = 90^0$. If $tan \angle ABC = 3.6$, then $AC : CD$ is

- $2BC : CE$
- $BC : 2CE$
- $CE : 2BC$
- $2CE : BC$

#### Problem 8.

$AB$ is a chord to a circle and $TAP$ is the tangent to the circle at $A$. If $\angle BAT = 75^0$ and $\angle BAC= 45^0$, where $C$ is a point on the circle, then $\angle ABC$ is,

- $40^0$
- $60^0$
- $70^0$
- $45^0$

#### Problem 9.

If radii of two circles be 6cm and 3cm and the length of the transverse common tangent be 8cm, then the distance between the centers is,

- $\sqrt {140}$cm
- $\sqrt {145}$cm
- $\sqrt {135}$cm
- $\sqrt {150}$cm

#### Problem 10.

All sides of a quadrilateral $ABCD$ touch a circle. If $AB=6$cm, $BC=7.5$cm and $CD=3$cm, then the length of $AD$ is,

- 3.5cm
- 2.5cm
- 1.5cm
- 4.5cm

### Answers to the exercise problems

**Problem 1: **c: 18cm.

**Problem 2: **b: 36.

**Problem 3: **c: 1 : 4.

**Problem 4: **d: $108^0$.

**Problem 5: **b: 2.

**Problem 6: **a: $2\sqrt{5}$cm.

**Problem 7: **a: $2BC : CE$.

**Problem 8: **b: $60^0$.

**Problem 9: **b: $\sqrt {145}$cm.

**Problem 10: **c: 1.5cm.

The detailed solutions to these questions are available in * SSC CGL level Solution Set 21 on Geometry 3*.

### Guided help on Suresolv Geometry

All of Suresolv Geometry articles are listed with links at the end, but this is an *unguided* list and **may not be up-to-date.**

To use the *extensive range of articles on geometry* problem solving **with best results**, *follow the guide instead,*

**5 step Suresolv Geometry Reading and Practice Guide for SSC CGL Tier II and Other competitive exams.**

*Basically, it is how to read and practice Suresolv Geometry guide.*

**It contains high school math articles on Geometry and even list of puzzles on Geometry.**

**The guide list of articles will be always UPTODATE.**

Wish you all the sure success.

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**Basic and rich Geometry concepts part 7, Laws of sines and cosines**

**Basic and rich Geometry concepts part 6, proof of triangle area from medians**

**Basic and rich Geometry concepts part 5, proof of median relations**

**Basic and rich Geometry concepts part 4, proof of arc angle subtending concept**

**Geometry, basic and rich concepts part 3, Circles**

**Geometry, basic concepts part 2, Quadrilaterals polygons and squares**

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**SSC CGL level Question Set 37, Geometry 5**

**SSC CGL level Solution Set 36, Geometry 4**

**SSC CGL level Question Set 36, Geometry 4**

**SSC CGL level Solution Set 21, Geometry 3**

**SSC CGL Level Question Set 21, Geometry 3**

**SSC CGL level Solution Set 20, Geometry 2**

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**SSC CGL level Solution Set 18, Geometry 1**

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