Question: Find the equation of the circle passing through the following points:

Let (h,k) be the centre of the circle and r be the radius. So, (x - h)² + (y - k)² = r² is the required equation of the circle.


a) (0,0), (2,0) and (0,4)
Solution:

Since, the circle passes through (0,0), (2,0) and (0,4), we get,
i) (0 - h)² + (0 - k)² = r²
ii) (2 - h)² + (0 - k)² = r²
iii) (0 - h)² + (4 - k)² = r²

Put value of r² from equation (ii) in equation (i), we get,
(0 - h)² + (0 - k)² = (2 - h)² + (0 - k)²
or, h² + k² = (4 - 4h +h²) + k²
or, h² - h² + 4h = 4
or, 4h = 4
So, h = 1

Put value of r² from equation (iii) in equation (i), we get,
(0 - h)² + (0 - k)² = (0 - h)² + (4 - k)²
or, h² + k² = h² + (16 - 8k + k²)
or, 8k + k² - k² = 16
or, 8k = 16
So, k = 2

So, centre (h,k) = (1,2)

Take (x,y) = (2,0) and (h,k) = (1,2), we have,
(2 - 1)² + (0 - 2)² = r²
or, 1 + 4 = r²
So, r = √5

And,
Equation of circle when centre (h,k) = (1,2) and radius (r) = √5 is
(x - h)² + (y - k)² = r²
or, (x - 1)² + (y -2)² = (√5)²
or, x² - 2x + 1 + y² - 4y +4 = 5
or, x² + y² -2x -4y = 0 is the required equation.

Hence, the required equation of the given circle is x² + y² - 2x - 4y = 0.


a) (1,1), (4,4) and (5,1)
Solution:

Since, the circle passes through (0,0), (2,0) and (0,4), we get,
i) ( - h)² + (1 - k)² = r²
ii) (4 - h)² + (4 - k)² = r²
iii) (5 - h)² + (1 - k)² = r²

Put value of r² from equation (ii) in equation (i), we get,
(1 - h)² + (1 - k)² = (4 - h)² + (4 - k)²

or, (1 -2h +h²) + (1 - 2k +k²) = (16 - 8h +h²) + (16 - 8k + k²)

or, 6h + 6k = 30
or, 6(h + k) = 30
or, h + k = 5
or, h = 5 - k --- (iv)


Put value of r² from equation (iii) in equation (i), we get,
(1 - h)² + (1 - k)² = (5 - h)² + (1 - k)²

or, (1- h)² - (5 - h)² = (1 - k)² - (1 - k)²

or, (1 - 2h + h²) - (25 - 10h + h²) = 0

or, 8h - 24 = 0
or, 8h = 24
So, h = 3

Put value of h = 3 in equation (iv), we get,
h = 5 - 3, h = 2

So, centre (h,k) = (3,2)

Take (x,y) = (1,1) and (h,k) = (3,2), we have,
(1 - 3)² + (1 - 2)² = r²
or, 4 + 1 = r²
So, r = √5

And,
Equation of circle when centre (h,k) = (3,2) and radius (r) = √5 is
(x - h)² + (y - k)² = r²
or, (x - 3)² + (y -2)² = (√5)²
or, x² - 6x + 9 + y² - 4y +4 = 5
or, x² + y² -6x -4y + 8 = 0 is the required equation.

Hence, the required equation of the given circle is x² + y² - 6x - 4y + 8 = 0.


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