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

Calculation:

Given,

The circle’s equation is

\(x^{2} + y^{2} + 2x + 2y + 1 = 0\)

Rewrite by completing squares:

\((x + 1)^{2} + (y + 1)^{2} = 1\)

∴ center = (-1, -1), radius r = 1

For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then

\(a\sqrt{2} = 2 \;\Longrightarrow\; a = \sqrt{2}.\)

Each vertex lies a half‐side \(= \tfrac{a}{2} = \tfrac{1}{\sqrt{2}} \) from the center along both axes. Since the center is ( -1, -1), the four vertices are

\(\displaystyle \bigl(-1 \pm \tfrac{1}{\sqrt{2}},\; -1 \pm \tfrac{1}{\sqrt{2}}\bigr).\)

One of these vertices is

\(\bigl(-1 + \tfrac{1}{\sqrt{2}},\; -1 - \tfrac{1}{\sqrt{2}}\bigr).\)

Hence, the correct answer is Option 3

Calculation:

Given two diameters of the circle are along the lines x + y = 0 and x - y = 0. The center of the circle is at their intersection.

We have, 

\(x - y = 0 \;\Longrightarrow\; x = y.\)

Substitute into x + y = 0:

\(x + x = 0 \;\Longrightarrow\; 2x = 0 \;\Longrightarrow\; x = 0,\; y = 0.\)

Thus, the center is (0, 0).

Also radius,

\(r = \frac{10}{2} = 5.\)

the equation of the circle with center (0,0) and radius 5 is

\(x^{2} + y^{2} = 5^{2} = 25.\)

Hence, the correct answer is Option 2.

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.

Explanation:

Given

Equation of circle is  

(x2 - 4x + 3) + (y2 - 6y + 8) = 0

⇒ (x – 3) (x – 1) + (y – 4) (y – 2)= 0

So the possible ends of the diameters are (3, 4), (3, 2), (1, 4) and (1, 2). 

Also, Radius = √2 

center = (2,3)

So the required pair of ends of diameters are (I) (1, 2) and (3, 4) and (II) (1, 4) and (3, 2).

∴ Option (a) is correct.

Given:

Centre of circle (h, k) = (−3, 2)

Area of circle = 176 units^2 .

Formula Used:

The area of the circle = πr2

Where, the radius r of the circle.

The equation of the circle in standard form

(x - h)2 + (y - k)2 = r2

Where, (h,k) is the center of the circle.

Calculation:

⇒ 176 = πr²

⇒ r² = 176 / π

So the equation of the circle is

(x + 3)² + (y - 2)² = 176 / π

Use, π = 22/7

(x + 3)² + (y - 2)² = 176 × (7/22)

∴ (x + 3)² + (y - 2)² = 56.

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:

If (x1, y1) and (x2, y2) are the end points of the diameter of a circle. Then the equation of such a circle is (x – x1) ⋅ (x – x2) + (y – y1) (y – y2) = 0

CALCULATION:

Given: The end points of the diameter of a circle are A (3, 2) and B (2, 5).

As we know that, if (x1, y1) and (x2, y2) are the end points of the diameter of a circle then the equation of such a circle is (x – x1) ⋅ (x – x2) + (y – y1) (y – y2) = 0

Here, x1 = 3, y1 = 2, x2 = 2 and y2 = 5

⇒ (x - 3) ⋅ (x - 2) + (y - 2) ⋅ (y - 5) = 0

⇒ x2 + y2 - 5x - 7y + 16 = 0

So, the equation of the required circle is: x2 + y2 - 5x - 7y + 16 = 0

Hence, option C is the correct answer.

Concept:

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

Calculation:

Using the formula for the equation of a circle with a given center and radius, we can write the equation as:

(x - 2)² + (y + 3)² = 5²

⇒ (x² - 4x + 4) + (y² + 6y + 9) = 25

⇒ x2 + y² - 4x + 6y - 12 = 0.

Concept:

The equation of circle with centre at (h, k) and radius 'r' is 

(x - h)2 + (y - k)2 = r2

Calculation:

We know that, the equation of circle with centre at (h, k) and radius 'r' is 

(x - h)² + (y - k)² = r²

 Here, centre (h, k) = (2, - 3) and radius r = 5 units.

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is 

(x - 2)2 + (y + 3)2 = 52

⇒\(\rm x^2 - 4x + 4 + y^2 +6y +9 = 25\)

⇒\(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \)

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is \(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \).

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 a circle whose centre is (2, - 1) and which passes through the point (3, 6)

Let the radius of the required circle be r units

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 = - 1

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

∵ The circle passes through the point (3, 6)

So, x = 3 and y = 6 will satisfy the equation (1)

⇒ (3 - 2)² + (6 + 1)² = r²

⇒ r² = 50

So, the equation of the required circle is (x - 2)2 + (y + 1)2 = 50

⇒ x² + y² - 4x + 2y - 45 = 0

So, the required equation of circle is x2 + y2 - 4x + 2y - 45 = 0

Hence, option D is the correct answer.

Concept:

Equation of circle having centre (h, k) and radius r is (x – h)² + (y – k)² = r²

Calculation:

Here, circle touching both the axes. Let it touches at (a, 0) and (0, a), so centre will be (a, a)

Now, Radius = a = 5

So, centre = (5, 5) and radius = 5

Now, Equation of circle having centre (5, 5) and radius 5

(x - 5)² + (y - 5)² = 5²

⇒ x² + 25 - 10x + y² + 25 - 10y = 25

⇒ x² + y² - 10x - 10y + 25 = 0

Hence, option (2) is correct. 

Concept:

Standard equation of a circle:

\(\rm (x-h)^2+(y-k)^2=R^2\)

Where centre is (h, k) and radius is R.

Note: The intersection of the diameters is the centre of the circle.

Distance between a point on a circle and the centre is the radius of the circle.

Distance between 2 points (x1, y1) and (x2, y2) is:

D = \(\rm \sqrt{(y_2-y_1)^2 + (x_2 -x_1)^2}\)

Perpendicular distance of a point (x₁, y₁) from the line ax + by + c = 0

D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

Calculation:

Given the circle touches the x and y axis, i.e., x and y axis are tangent to the circle.

Equation of the x-axis is y = 0

Radius is perpendicular distance of centre (-2, -2) from the tangent (y = 0) 

Radius r = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

⇒ r = \(\rm \left|0\times(-2)+1\times(-2)+0\over\sqrt{0^2+1^2}\right|\)

⇒ r = |-2| = 2

Equation of circle having centre (-2, -2) and radius r = 2 is:

\(\rm (x-h)^2+(y-k)^2=R^2\)

⇒ (x-(-2))² + (y-(-2))² = 2² 

⇒ x² + y² + 4x + 4y + 8 = 4

⇒ x2 + y2 + 4x + 4y + 4 = 0

Concept:

The general form of the equation of a circle is:

x² + y² + 2gx + 2fy + c = 0

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

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

Calculation:

We have, x2 + y2  - 4x - 4y + 4 = 0

On comparing it with the general equation of the circle, we get,

g = -2, f = -2 and c = 4

∴ Radius of the circle = \(\sqrt{g^{2}+f^{2}-c}\)

\(⇒ \sqrt{(-2)^{2}+(-2)^{2}-4}\)

\(⇒ \sqrt{4+4-4}\)

⇒ 2

Hence, the radius of the circle is 2.

Concept: 

Equation of circle having centre (h, k) and radius r is 

 (x – h) ² + (y – k) ² = r²  

Calculation: 

Given, centre of circle ( 3, 2) and radius 7 units

We know that , equation of circle is  (x – h) 2 + (y – k) 2 = r2   

According to the question, the equation of circle is 

( x - 3 )² + ( y - 2 )² = 7² 

⇒ x²+ 9 - 6x + y²+ 4 - 4y = 49 

⇒ x² + y²- 6x - 4y - 36 = 0. 

∴ The correct option is 2. 

Concept: 

If Circle with centre (h, k) and passes through an arbitrary point P (x, y) on the circle, then the radius of circle, 

| CP | = r = \(\rm \sqrt{\left ( x-h \right )^{2}+\left ( y-k \right )^{2}}\)  

Equation of circle having centre (h, k) and radius r is 

(x – h) 2 + (y – k) 2 = r2   

Calculation:  

Let C (2, -3) be the centre of of the given circle and let it passe through the point P (3, 4). Then, the radius of circle 

| CP | = r = \(\rm \sqrt{\left ( 3-2 \right )^{2}+\left ( 4+3 \right )^{2}}\) = \(\sqrt{50}\) 

∴ Required equation of circle is , 

( x - 2 )2 + ( y + 3 )2 =  ( \(\sqrt{50}\) )² 

⇒ x² + y² - 4x + 6y - 37 = 0. 

The correct option is 4.