Elements of a Circle MCQ Quiz - Objective Question with Answer for Elements of a Circle - Download Free PDF

Given:

2x 2 + 2y2 − 8x − 12y − 24 = 0

Formula Used:

The general equation of a circle 

(x - h)2 + (y - k)2 = r2

Where center of circle = (h, k) & radius = r

Calculation:

⇒ 2x2 − 8x + 2y² −12y − 24 = 0

⇒ x2 − 4x + y2 − 6y − 12 = 0

⇒ x2 − 4x + 4 - 4 + y2 − 6y + 9 - 9 - 12 = 0

⇒ (x - 2)2 + (y - 3)2 - 25 = 0

⇒ (x - 2)2 + (y - 3)2 = 25

⇒ (x - 2)2 + (y - 3)2 = r²

∴ Center = (2, 3) and radius = 5 

Concept:

Inscribed Square in a Circle Inside an Equilateral Triangle:

• In an equilateral triangle of side length a, the radius r of the inscribed circle (incircle) is given by:
• r = a × √3 / 6
• This formula is derived using the relationship between the area of the triangle and its semiperimeter.
• The diagonal of a square inscribed in a circle is equal to the diameter of the circle.
• If the diagonal of a square is d, then its side length is d / √2, and area is (d / √2)² = d² / 2.

Calculation:

Given, side of equilateral triangle = a

⇒ Radius of inscribed circle = r = a × √3 / 6

⇒ Diameter of the circle = 2r = 2 × (a × √3 / 6) = a × √3 / 3

⇒ Diagonal of square = a × √3 / 3

Now, side of square = (diagonal) / √2

⇒ Side = (a × √3 / 3) / √2 = a × √3 / (3√2)

⇒ Area = side² = (a × √3 / (3√2))²

⇒ Area = a² × 3 / (9 × 2) = a² / 6

∴ Hence, the area of the inscribed square is a² / 6.

Concept:

General Equation of a Circle

⇒ \( (x - h)^2 + (y - k)^2 = r^2 \)

where (h , k) is the center and r is the radius of the circle

Calculation:

Given Conditions for the Center(centre lies on a line 2x + 3y = 3)

⇒ \( 2h + 3k = 3 \)      ---------(1)

Since circle  passes through the points (3, 0) and (0, -2) : 

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⇒\( (3 - h)^2 + k^2 = h^2 + (-2 - k)^2 \)

⇒\( 6h + 4k = 5 \)     --------(2)

Solving equation 1 and 2

⇒  \( 2h + 3k = 3 \\ 6h + 4k = 5 \)

⇒  \( h = \frac{3}{10}, \quad k = \frac{4}{5} \)

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⇒  \( r^2 = \left( 3 - \frac{3}{10} \right)^2 + \left( \frac{4}{5} \right)^2 \)

⇒  \( r^2 = \frac{793}{100} \)

Now, 

⇒  \( \left( x - \frac{3}{10} \right)^2 + \left( y - \frac{4}{5} \right)^2 = \frac{793}{100} \)  

⇒ \(​\frac{(10x-3)^2} {100} + \frac{(5y-4)^2}{25} = \frac{793}{100}\) 

⇒ \(​\frac{(10x-3)^2} {100} + 4\frac{(5y-4)^2}{100} = \frac{793}{100}\) 

⇒ \( (10x-3)^2 + (10y - 8)^2 = 793 \)

⇒ \( 100x^2 + 9 - 60x + 100y^2 + 64 -160 y = 793 \)

⇒ \( 100 x^2 + 100 y^2 - 60 x -160 y = 720 \)

Divide  by 10 and simplifying

⇒\( 10x^2 + 10y^2 - 6x - 16y - 72 = 0 \)

Hence option 2 is true

A pie diagram (or pie chart) is a circular statistical graphic that represents data as slices of a circle. Each slice or sector represents a proportion of the total dataset.

• The size of each slice is proportional to the quantity it represents. To construct a pie diagram, it is essential to understand the mathematical basis for the circle's area, as this area will determine how the data is visualized.

Key Points

• The area of a circle is calculated by the formula πr², where r is the radius of the circle.
• This formula is used to calculate the total area of the circle, which is then divided into sections that represent different proportions of the total data in a pie chart.

Hint

• 2πr is the formula for the circumference of the circle, not the area. It calculates the distance around the circle.
• 3πr does not correspond to any standard formula in geometry and is not used to calculate either the area or circumference of a circle.
• π(r/2) would not give the correct area of the circle. Instead, it seems to be a misrepresentation of the formula for the radius of a circle or some other calculation that is not related to the area of the circle.

Hence, the correct answer is πr².

Concept Used:

1. The equation of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r².

2. If the center lies on the y-axis, then h = 0.

Calculation

Given: Center lies on the y-axis, so the center is (0, k).

The circle passes through (-4, 3) and (3, -4).

Equation of the circle: (x - 0)² + (y - k)² = r²

⇒ x² + (y - k)² = r²

Substitute (-4, 3):

(-4)² + (3 - k)² = r²

⇒ 16 + 9 - 6k + k² = r²

⇒ 25 - 6k + k² = r² ...(1)

Substitute (3, -4):

(3)² + (-4 - k)² = r²

⇒ 9 + 16 + 8k + k² = r²

⇒ 25 + 8k + k² = r² ...(2)

From (1) and (2):

25 - 6k + k² = 25 + 8k + k²

⇒ -6k = 8k

⇒ 14k = 0

⇒ k = 0

Substitute k = 0 into (1):

25 - 6(0) + 0² = r²

⇒ r² = 25

⇒ r = 5

∴ The radius of the circle is 5.

The correct option is 4) 5.

Given:

∠ACB = 116°

Concept Used:

When a quadrilateral is inscribed in a circle, the opposite angles of it are supplementary angles.

The angle subtended at the center is always double the angle subtended at the remaining arc.

The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

Calculation:

Point P is taken on the major arc of the circle.

Then, A & P, B & P, C & B, and A & C are joined.

From cyclic quadrilateral APBC

∠ACB = 116°

Now, ∠APB = (180 – 116)° = 64°

Now, ∠AOB = (64 × 2)°

⇒ 128°

Since OA = OB = radius of the circle

So, from quadrilateral AOBD

∠ADB = 360° - (∠OBD + ∠OAD + ∠AOB)

⇒ ∠ADB = 360° - (90° + 90° + 128°)

⇒ ∠ADB = 360° - 308°

⇒ ⇒ ∠ADB = 52° 

∴ The required measure of ∠ADB is 52°.

Concept:

The general second degree equation of a circle in x and y is given by:  x2 + y2 + 2gx + 2fy + c = 0 with centre (-g, -f) and radius

 \(\rm r = \sqrt {{g^2} + {f^2} - c} \)

Calculation:

Given:   2x2 + 2y2 + 8x + 8y + 4 = 0 

⇒ 2 × (x2 + y2 + 4x + 4y + 2) = 0

⇒ x2 + y2 + 4x + 4y + 2 = 0 are equation of circle with centres C and radius r 

By comparing the equation of the circle with the equation  x 2 + y2 + 2gx + 2fy + c = 0 we get 

g = 2, f =  2 and c = 2

As we know, radius = \(\rm r = \sqrt {{g^2} + {f^2} - c} \)

\(\rm r = \sqrt {{4} + {4} - 2} \)

r = √6 unit

Concept: 

Any line from the center that bisects a chord is perpendicular to the chord.

Pythagoras theorem: 

h² = b² + p²

Calculation:

r² = 4² + 3²

⇒ r = 5

∴ The diameter of the circle = 2r = 2 × 5

⇒ 10 cm 

Shortcut TrickBy triplets:

3, 4 and 5 

The length of the diameter = 2× 5 = 10

Concept:

Let x2 + y2 = r² is the equation of circle. then (0, 0) is the origin and r is the radius of the circle. 

Calculations:

We know that, x2 + y2 = r2 is the equation of circle. then (0, 0) is the origin and r is the radius of the circle.

Given equation of circle is x² + y² + x + c = 0, which is passing through the origin.

i.e. c = 0.

⇒x² + y² + x = 0.

⇒ x² + x + \(\frac{1}{4}\)- \(\frac{1}{4}\) + y² = 0

⇒x² + x + \(\frac{1}{4}\)+ y² = \(\frac{1}{4}\)

⇒\(\left ( x+\frac{1}{2} \right )^2 +y^2 = \left (\frac{1}{2} \right )^2\)

which is equation of circle with radius is \(\frac{1}{2}\).

Hence, the radius of the circle x2 + y2 + x + c = 0 passing through the origin is \(\frac{1}{2}\).

Concept:

The general form of the equation of a circle is:

x2 + y2 + 2gx + 2fy + c = 0

The centre of the circle is (-g, -f).

The radius of the circle is \(√{g^{2}+f^{2}-c}\).

Calculation:

Given equation of a circle is, 

4x2 + 4y2 - 20x + 12y - 15 = 0

x2 + y2 - 5x + 3y - 15/4 = 0

On comparing from the general equation of circle

2g = 5, or g = 5/2

2f = 3 or f = 3/2 & c = -15/4

By using the above formula

Radius = \(√{\frac{25}{4}+\frac{9}{4}-(-\frac{15}{4})}\)

Radius = \(√{\frac{25+9+15}{4}} =√ \frac{49}{4} \) = √12.25

Radius = 3.5

Alternate Method

Given equation of a circle is, 

4x2 + 4y2 - 20x + 12y - 15 = 0

⇒ [4x2 - 20x] + [4y2 + 12y] - 15 = 0

⇒ [(2x)² - 2 . 2x . 5 + 5² - 5²] + [(2y)2 + 2 . 2y . 3 + 32 - 32] - 15 = 0

⇒ [(2x - 5)2 - 52] + [(2y + 3)2 - 32] - 15 = 0

⇒ (2x - 5)2  + (2y + 3)2 - 15 - 25 - 9 = 0

⇒ (2x - 5)² + (2y +3)² = 49

⇒ 4(x - \(5\over2\))² + 4(y + \(3\over2\))² = 49

⇒ (x - \(5\over2\))2 + (y + \(3\over2\))² = (\(7\over2\))²

∴ The radius of the circle = \(7\over2\)units = 3.5 units

Concept:

The distance between a point P(x₁, y₁) and the line ax + by + c = 0 is given by:

Distance = \(\rm \dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\).

Calculation:

Since the circle touches the line at a point, the radius of the circle will be the distance between the center of the circle and the given line.

Using the formula for the distance between a point (3, 4) and a line (3x + 4y - 5 = 0), we get:

Radius = \(\rm \dfrac{|3(3)+4(4)-5|}{\sqrt{3^2+4^2}}\) = \(\rm \dfrac{20}{5}\) = 4.

Concept:

The general second degree equation of a circle in x and y is given by:  x2 + y2 + 2gx + 2fy + c = 0 with centre (-g, -f) and radius

 \(\rm r = \sqrt {{g^2} + {f^2} - c} \)

Calculation:

Given:   2x2 + 2y2 + 24x + 24y + 8 = 0 

⇒ 2 (x2 + y2 + 12x + 12y + 4) = 0

⇒ x2 + y2 + 12x + 12y + 4 = 0 are equation of circle with centres C and radius r 

By comparing the equation of the circle with the equation  x 2 + y2 + 2gx + 2fy + c = 0 we get 

g = 6, f =  6 and c = 4

As we know, radius = \(\rm r = \sqrt {{g^2} + {f^2} - c} \)

\(\rm r = \sqrt {{36} + {36} - 4} \)

r = 2√17 unit

∴ The radius of circle is 2√17

Concept:

Length of arc = \(\frac{\theta }{{360}} \times 2\pi r\)

Calculation:

Given Data;

Diameter = 44 cm

\({\rm{Radius}} = {\rm{}}\frac{{{\rm{Diameter}}}}{2}{\rm{}} = {\rm{}}\frac{{44}}{2}{\rm{}} = {\rm{\;}}22{\rm{cm}}\)

Chord length = 22 cm

Figure

In triangle AOB;

OA = OB = Radius = 22 cm

AB = 22 cm

∵ All sides are equal so triangle is equilateral

∴ All angles are same, \(\theta = 60\)

\({\rm{Length\;of\;arc}} = {\rm{}}\frac{{\rm{\theta }}}{{360}} \times 2{\rm{\pi r}} = {\rm{}}\frac{{60}}{{360}} \times 2 \times \frac{{22}}{7} \times 22{\rm{}} = {\rm{}}\frac{1}{6} \times \frac{{968}}{7}{\rm{}} = {\rm{}}\frac{{484}}{{21}}{\rm{\;cm}}\)

Concept: 

General form of the equation of a circle, x2 + y2 + 2gx + 2fy + c = 0 

Radius = \(\rm\sqrt{g^{2}+f^{2}-c}\)  

Calculation: 

The given equation of circle is , 3x2+ 3y2- 6x+ 12y- 13= 0 

⇒ x²+ y²- 2x + 4y - \(\frac{13}{3}\) = 0         ....(i)

On compare with standard equation of circle x2 + y2 + 2gx + 2fy + c = 0  

where , g = 1 , f = 2 and c = - \(\frac{13}{3}\)  

We know that , radius of circle = \(\rm\sqrt{g^{2}+f^{2}-c}\) 

⇒ radius = \(\sqrt{1^{2}+2^{2}-\left ( -\frac{13}{3} \right )}\) 

⇒ radius = \(\sqrt{\frac{28}{3}}\) units. 

The correct option is 1.

Given:

Area of circle = Perimeter of the circle

Formula Used:

Area of circle = πr²

Perimeter of the circle = 2πr

Calculation:

According to question,

⇒ πr² = 2πr

⇒ r = 2 unit

∴ The radius of the circle is 2 unit.