CRE 3 - 2 Circles | Geometry - Circles

Answer: Option A

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Explanation :

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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).

Answer: 8

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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.

Answer: Option D

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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).

Answer: Option A

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Explanation :

Given, m ∠CDF = 140°

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∴ 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).

Answer: Option B

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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 × 1² - $\frac{\sqrt{3}}{4}$ × 1²

⇒ 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).

Answer: Option A

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Explanation :

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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).

Answer: 7.2

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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.

Answer: Option B

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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.

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

⇒ AB² = AC² + BC²

⇒ 26² = (6+ 4)² + BC²

⇒ BC² = 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).