Circles MCQ Quiz - Objective Question with Answer for Circles - Download Free PDF

Calculation: 

m₁m₂ = – 1 so right angle equation circle is

⇒ (x – 4) (x – 0) + (y – 6) (y – 0) = 0

⇒ x² + y² – 4x – 6y = 0

⇒ (k,3k) lies on it so

⇒ k² + 9k² – 4k – 18k = 0

⇒ 10k² – 22k = 0 

⇒ k = 0, \(\frac{11}{5}\)

k = 0 is not possible so k = \(\frac{11}{5}\)

also r = \(\sqrt{4+9}=\sqrt{13}\)

so 10k + r² = \(\text { 10. } \frac{11}{5}+(\sqrt{13})^{2}=35\)

Hence, the Correct answer is Option 4.

Concept:

Circle Equation and Radius:

• The radius of a circle touching a line can be found using the perpendicular distance from the center to the line.
• The circle equation is \( (x - a)^2 + (y - b)^2 = r^2 \), where \( (a,b) \) is the center and \( r \) is the radius.

Calculation:

Center of circle is \( (a, 0) \).

Radius \( r \) is distance from center to line \( x + y = 3 \):

\( r = \left| \frac{a - 0 - 3}{\sqrt{2}} \right| = \left| \frac{a - 3}{\sqrt{2}} \right| \)

Equation of circle:

\( (x - a)^2 + y^2 = \left( \frac{a - 3}{\sqrt{2}} \right)^2 \)

Circle passes through point \( (-9,4) \):

\( (-9 - a)^2 + 4^2 = \frac{(a - 3)^2}{2} \)

Expanding and simplifying:

\( (a + 9)^2 + 16 = \frac{(a - 3)^2}{2} \)

\( 2(a^2 + 18a + 81 + 8) = a^2 - 6a + 9 \)

\( 2a^2 + 36a + 194 = a^2 - 6a + 9 \)

\( a^2 + 42a + 185 = 0 \)

Factorizing:

\( (a + 37)(a + 5) = 0 \implies a = -37, -5 \)

Calculating radii:

\( r_1 = \left| \frac{-37 - 3}{\sqrt{2}} \right| = \frac{40}{\sqrt{2}} = 20\sqrt{2} \)

\( r_2 = \left| \frac{-5 - 3}{\sqrt{2}} \right| = \frac{8}{\sqrt{2}} = 4\sqrt{2} \)

Absolute difference between squares of radii:

\( |r_1^2 - r_2^2| = |(20\sqrt{2})^2 - (4\sqrt{2})^2| = |800 - 32| = 768 \)

Hence, the correct answer is 768.

Ans. (1)

Sol.

Equation of circle

x² + (y – (6 – r))² = r²

touches \(\sqrt3\)x - y = 0

p = r 

\(\frac{|0-(6-r)|}{2}=r\)

|r – 6| = 2r

r = 2 

∴ Circle x² + (y – 4)² = 4

(2, 4) Satisfies this equation

Explanation:

Given, C₁: x2 + y2 + 2kx + 4y - 4 = 0

C₂: x2 + y2 + 6x - 2y + 6 = 0

Center of the C₁ is (- k,  - 2) and r₁ = \(\sqrt{k^2 + 4 + 4}\)  = \(\sqrt{k^2 +8}\)

According to the question, the center is in the 4th quadrant. So,  k < 0  (∵ In the 4^th quadrant x - coordinate is positive.)

and the center of C₂ is ( - 3, 1) and r₂ = √9 + 1 - 6 = 2 

It is given that circle C₁ touches the circle C₂. Therefore, the distance between the center of the circles is equal to the sum of their radii.

C₁C₂ = r₁ + r₂

⇒ \(\sqrt{(-3 +k)^2 +3^2}\) = \(\sqrt{k^2 + 8 } + 2\) 

⇒ \(\sqrt{(k^2 -6k + 9 + 9 )}\)  = \(\sqrt{k^2 + 8 } +2\)

⇒ \(\sqrt{(k^2 -6k + 18 )}\) = \(\sqrt{k^2 + 8 } +2\)

Squaring on both the sides, we get

⇒ k2 - 6k + 18 = k2 + 8 +  4 + 4 \(\sqrt{k^2 + 8 } \)

⇒ - 6k + 6 =  4 \(\sqrt{k^2 + 8 } \)

⇒ - 3k + 3 = 2 \(\sqrt{k^2 + 8 } \)

Again squaring on both the sides, we get

9k² - 18k + 9 = 4 (k² + 8)

9k² - 18k + 9 = 4k² + 32

5k² - 18k - 23 = 0 

 k =  - 1, \(\frac{23}{5}\)

∵ k < 0, therefore  k  = - 1

Concept:

The equation of circle with centre at (a, b) and radius r is given by : (x - a)2 + (y - b)2 = r2 

Calculation:

Let equation of circle be (x - a)2 + (y - b)2 = r2 

Given the centre is (1, –2)

⇒ Equation of circle becomes  (x – 1)2 + (y + 2)2 = r2 ...(i)

Now, (i) passes through the intersection of lines  3x + y = 14 and 2x + 5y = 18.

Solving the two equations, we get x = 4 and y = 2

Putting the value of x and y in (ii), we get:

⇒ (4 – 1)2 + (2 + 2)2 = r2 

⇒ 3² + 4² = r² 

⇒ r² = 25

Putting the value of r2 in (ii), we get:

(x – 1)2 + (y + 2)2 = 25

⇒ x² + 1 – 2x + y² + 4 + 4y = 25

⇒ x2 + y2 – 2x + 4y – 20 = 0

∴ Required equation of circle is x2 + y2 – 2x + 4y – 20 = 0.

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:

On X- axis y-intercept will be zero, similarly, on Y-axis x-intercept will be zero.

Calculation:

On Y-axis x-intercept = 0

So, 

\(\rm x^2+y^2 +4x-7y+12=0\\ ⇒ y^2-7y+12=0\\ ⇒ y^2-4y-3y+12=0\\ ⇒ y(y-4)-3(y-4) =0 \\ ⇒ (y-3)(y-4)=0\)

 y = 3 and 4

Points (0, 3) and (0, 4)

Now length =

\(=\sqrt{0^2+(3-4)^2}\\ =1\)

Hence, option (4) is correct.

Given:

∠CAB = 52° and ∠ABD = 47°

Concept used:

The angle subtended by the diameter at any point on the circumference is equal to 90°

Calculation:

According to the concept,

∠ACB = ∠ADB = 90°

So, ∠CBA = 180° - (90° + 52°)

⇒ 180° - 142°

⇒ 38°

Similarly,

∠BAD = 180° - (90° + 47°)

⇒ 180° - 137°

⇒ 43°

So, ∠CAD = 52° + 43°

⇒ 95°

∠CBD = 38° + 47°

⇒ 85°

Now,

∠CAD - ∠CBD = 95° - 85°

⇒ 10°

∴ The difference (in degrees) between the measures of ∠CAD and ∠CBD is 10.

Given:

Radius + Diameter = 84 cm

Formula used:

Diameter = 2 x Radius

Circumference = 2πr

Calculations:

Let the radius be R then, Diameter be 2R.

Now, according to question, 

R + 2R = 84

⇒ 3R = 84

⇒ R = 28 

So, The circumference = 2πr = 2 × \(\frac{22}{7}\) × 28 = 176 cm

Now, difference between Circumference and radius is

= 176 - 28

= 148 cm

Hence, the required difference is 148 cm.

Concept:

The length of the tangent from an external point (x₁, y₁) to the circle x² + y² = a² is \(\sqrt {x_1^2 + y_1^2 - {a^2}}\)

Calculation:

Given: Equation of circle x² + y² = 9 and the point (4, 0).

As we know that, the length of the tangent from an external point (x₁, y₁) to the circle x² + y² = a² is \(\sqrt {x_1^2 + y_1^2 - {a^2}}\)

Here, x₁ = 4 , y₁ = 0 and a² = 9.

CONCEPT:

Equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2

CALCULATION:

Here, we have to find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4

First let's find the point of intersection of the lines  3x + 2y = 11 and 2x + 3y = 4

So, by solving the equations  3x + 2y = 11 and 2x + 3y = 4, we get x = 5 and y = - 2

So, the required circle passes through the point (5, - 2)

Let the radius of the required circle be r

As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2

Here, we have h = 2 and k = - 3

⇒ (x - 2)² + (y + 3)² = r² ------------(1)

∵ The circle  passes through the point (5, - 2)

So, x = 5 and y = - 2 will satisfy the equation (1)

⇒ (5 - 2)² + (- 2 + 3)² = r²

⇒ r² = 10

So, the equation of the required circle is (x - 2)2 + (y + 3)2 = 10

⇒ x² + y² - 4x + 6y + 3 = 0

So, the equation of the required circle is x2 + y2 - 4x + 6y + 3 = 0

Hence, option A is the correct answer.

Given

Centre point are  (1, -2)

Radius = 4cm

Formula used

(x -a)² + (y - b)² = r²

where, a and b are point on centre

r = radius

x and y be any point on circle

Calculation

Put the value of  a, b and r in the formula

(x-1)² + (y + 2)²  = 16

⇒ x² + 1 - 2x + y² + 4 + 4y = 16

⇒ x² + y² - 2x + 4y = 11

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 \(\frac{1}{2}\).