Test: Circles: Equation of a Circle (8 July) - JEE MCQ

Given:
Diameter of a circle are (2, −1) and (−2, 2). 
Formula Used:
The equation of a circle whose end points of any diameter are (x₁, x₂) and (y₁, y₂) 
(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
Calculation:
(x - 2)(x + 2) + (y + 1)(y - 2) = 0
x² - 2x + 2x - 4 + y² - 2y + y - 2 = 0 
∴ x² + y²​ - y - 6 = 0

Given:
The circle passes through the point (4, 5)
The center of the circle is at (2, 2)
Formula Used:
The equation of circle with centre (a, b) and radius r is defined as
(x - a)² + (y - b)² = r²
Calculation:
Centre is at (2, 2) and point on the circle is (4, 5) 
Radius (r) = √[(4 - 2)² + (5 - 2)²]
⇒ r = √(4 + 9) = √13
Now, According to the formula used
Equation of circle = (x - 2)² + (y - 2)² = (√13)²
⇒ x² - 4x + 4 + y² - 4y + 4 = 13
⇒ x² + y² - 4x - 4y - 5 = 0
∴ The equation of circle is x² + y² - 4x - 4y - 5 = 0.

Concept:
Standard equation of a circle:
(x−h)²+(y−k)² = R²
Where centre is (h, k) and radius is R.
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:

Calculation:
Let coordinates of the centre are (h, k)
Given the circle passes through (2, 1), (1, 2) and (5, -6)
Now the radius r = 

Equating values of r form (i) and (ii)
⇒ -4h - 2k = -2h - 4k
⇒ h = k           ...(iv)
Equating values of r form (i) and (iii) 
⇒ -4h - 2k + 5 = -10h + 12k + 61
⇒ 6h - 14k = 56
As h = k,
⇒ -8k = 56
⇒ k = -7 = h 
∴ Coordinates of the centre are (-7, -7)
From equation (i)

⇒ r² = 81 + 64 = 145
∴ Equation of circle is:

⇒ (x + 7)² + (y + 7)² = 145

Concept: 
We know that the general equation for a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.
Solution:
Given: The circle equation is 

The general equation for a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius.
Hence, The center and radius of equation (1) is 
∴ The correct option is (3).

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

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

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.