Question: Find the equation of a circle which passes through the point (5,5) and (3,7) and centre lies on the line x - 4y = 1.

Solution:

In a circle,
Let the coordinates of the centre of the circle be (h,k)

According to the question,
Centre lies on the line
x - 4y = 1
or h - 4k = 1
or, h = 1 + 4k

Given, passing points (x,y) are (5,5) and (3,7).

Equation of the circle when (x - h)² + (y - k)² = r², we get,

(i) (5 - h)² + (5 - k)² = r²
(ii) (3 - h)² + (7 - k)² = r²

From equations (i) and (ii), we get,

or, (5 - h)² + (5 - k)² = (3 - h)² + (7 - k)²

or, 25 - 10h + h² + 25 - 10k + k² = 9 - 6h + h² + 49 - 14k + k²

or, 50 - 10h - 10k = 58 - 6h - 14k

or, 4k = 8 + 4h

or, k = 2 + h
[Put h = 1 + 4k from above]
or, k = 2 + 1 + 4k
or, -3 = 3k
So, k = -1

Put k = -1 in h= 1 + 4k, we get,
h = 1 + 4(-1)
= 1 - 4
= -3

So, centre (h,k) = (-3,-1)

When (x,y) = (5,5), centre (h,k) = (-3,-1), we get,
(x - h)² + (y - k)² = r²
or, {5 - (-3)}² + {5 - (-1)}² = r²
or, (5 +3)² + (5+1)² = r²
or, 64 + 36 = r²
or, 100 = r²
So, r = 10 units

When (h,k) = (-3,-1) and r = 10, equation of circle is given by
(x - h)² + (y - k)² = r²

or, {x - (-3)}² + {y - (-1)}² = 10²

or, (x +3)² + (y +1)² = 100

or, x² + 6x + 9 + y² + 2y + 1 = 100

or, x² + y² + 6x + 4y - 90 = 0 is the required equation.

Hence, the required equation of the given circle is x² +y² +6x +4y -90 = 0.

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