# CRE 3 - 2 Circles | Geometry - Circles

**CRE 3 - 2 Circles | Geometry - Circles**

One circle is drawn completely inside another circle such that no point is common to both the circles. How many common tangent(s) can be drawn?

- (a)
0

- (b)
1

- (c)
2

- (d)
Infinite

Answer: Option A

**Explanation** :

As shown in the figure, since one circle is completely inside the bigger circle such that no point is common to both the circles.

Hence, any tangent drawn to the inner circle always intersects the outer circle at two points thereby becoming a secant.

∴ No common tangent can be drawn for two circles, where one circle is completely inside the other and there is no point common to both the circles.

Hence, option (a).

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

Find the length of the Direct common tangent for two circles of radius 3 and 9 cm whose centers are 10 cm apart (in cm).

Answer: 8

**Explanation** :

We know the length of direct common tangent = $\sqrt{{d}^{2}-{\left({r}_{!}-{r}_{2}\right)}^{2}}$.

Hence, DCT = $\sqrt{{10}^{2}-{\left(9-3\right)}^{2}}$= $\sqrt{{10}^{2}-{6}^{2}}$ = 8.

Hence, 8.

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

In the previous question, what will be the length of the Transverse Common Tangent?

- (a)
12 cm

- (b)
5 cm

- (c)
10 cm

- (d)
Not possible

Answer: Option D

**Explanation** :

Here, since the distance between the two centers is less than the sum of the two radii, it means that the two circles intersect at two different points.

Hence, these two circles do not have a transverse common tangent.

Hence, option (d).

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

Two circles If m ∠CDF = 140°, what is m ∠BAG in the given figure? (Line ABC is the tangent at B and C to the circles with centers G and E respectively. Also, point F is the point where both circles are tangent to each other).

- (a)
10°

- (b)
20°

- (c)
30°

- (d)
40°

Answer: Option A

**Explanation** :

Given, m ∠CDF = 140°

∴ m ∠CHF = 180° - 140° = 40°.

Angles subtended by a chord in both segments are supplementary.

∴ m ∠CEF = 2 × m ∠CHF

∴ m ∠CEF = 2 × 40° = 80°

The angle subtended by the chord CF on the circumference at point D (which lies in the minor segment) of the circle measures half the angle subtended by the same chord at the center ‘E’.

Now, In Δ ACE,

m ∠ACE = 90° (because C is the point of contact of tangent AC and radius CE)

m ∠CEA + m ∠ACE + m ∠CAE = 180°

∴ m ∠CAE = 180° – (90° + 80°)

∴ m ∠CAE = m ∠BAG = 10°

Hence, option (a).

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

Two equal circles with centers A and B intersect at points P and Q, such that the center of each circle lies on the circumference of the other circle. The distance between centers of both the circles is 1 unit. What is the area of the common region between the two circles?

- (a)
$\frac{\mathrm{\pi}-\sqrt{3}}{4}$

- (b)
$\frac{2\mathrm{\pi}}{3}$ - $\frac{\sqrt{3}}{2}$

- (c)
π/2

- (d)
π/4

- (e)
None of these

Answer: Option B

**Explanation** :

We need to find the area of the shaded part in the figure below.

The area of the common region will be twice the area of segment PBQ (shaded)

In ∆APB, AP = BP = AB = 1 cm. Hence, ∆APB is an equilateral triangle.

∴ ∠PAB = 60° = ∠QAB

∴ ∠PAQ = 120°

Area of segment PBQ = Area of sector APBQ - Area of triangle APQ

Now, Area of triangle APQ = Area of triangle APB

⇒ Area of segment PBQ = $\frac{120}{360}$ × π × 1^{2} - $\frac{\sqrt{3}}{4}$ × 1^{2}

⇒ Area of segment PBQ = $\frac{\mathrm{\pi}}{3}$ - $\frac{\sqrt{3}}{4}$ =

∴ Area of region common to both circles = 2 × $\left(\frac{\mathrm{\pi}}{3}-\frac{\sqrt{3}}{4}\right)$ = $\frac{2\mathrm{\pi}}{3}$ - $\frac{\sqrt{3}}{2}$

Hence, option (b).

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

Two equal circles with centers A and B intersect at points P and Q, such that the center of each circle lies on the circumference of the other circle. The distance between centers of both the circles is 1 unit. What is the area of the rhombus APBQ (in sq. units)?

- (a)
√3/2

- (b)
√3/4

- (c)
1/2

- (d)
4

Answer: Option A

**Explanation** :

In ΔAPB, AP = AB = PB = 1

∴ ΔAPB is an equilateral triangle with side 1.

Similarly, ΔAQB is also an equilateral triangle with side 1.

∴ Area of the rhombus APBQ = A(ΔAPB) + A(ΔAQB)

= $\frac{\sqrt{3}}{4}\times {1}^{2}+\frac{\sqrt{3}}{4}\times {1}^{2}$

∴ Area of a rhombus APBQ = $\frac{\sqrt{3}}{2}$ square units

Hence, option (a).

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

Three balls are placed inside a cone such that each ball is in contact with the edge of the cone and the next ball. Find r.

Answer: 7.2

**Explanation** :

When three or more circles have two common tangents then the radii of these circles is in Geometric Progression.

∴ r, 12 and 20 are in G.P.

⇒ $\frac{r}{12}=\frac{12}{20}$

⇒ r = 144/20 = 7.2.

Hence, 7.2.

Workspace:

**CRE 3 - 2 Circles | Geometry - Circles**

Two circles, one with centre A and radius 4 cm and the other with centre B and radius 6 cm, touch a line at points P and Q respectively. A and B lie on opposite sides of the line PQ. If AB = 26 then find PQ (in cm).

- (a)
25

- (b)
24

- (c)
23

- (d)
22

- (e)
None of these

Answer: Option B

**Explanation** :

Let AP is extended while a line from B is drawn which is parallel to PQ. Both these lines meet at C as shown in the figure.

From the diagram above, triangle ABC is right angled at vertex C.

⇒ AB^{2} = AC^{2} + BC^{2}

⇒ 26^{2} = (6+ 4)^{2} + BC^{2}

⇒ BC^{2} = 676 - 100 = 576

⇒ BC = 24

Now, BC = PQ = 24 (Since PQBC is a rectangle)

**Alternately**,

PQ is the transverse common tangent

Lenth of transverse common tangent = $\sqrt{{\mathrm{d}}^{2}-{({\mathrm{r}}_{1}+{\mathrm{r}}_{2})}^{2}}$

⇒ PQ = $\sqrt{{26}^{2}-{(4+6)}^{2}}$ = 24

Hence, option (b).

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