Question: A circle has radius 5 units and centre is the point of intersection of the lines 2x - y = 5 and x - 3y +5 = 0. Find the equation of the circle.

Solution:
Given,
In a circle,
Radius (r) = 5 units
Centre (h,k) = point of intersection of (2x - y = 5) and (x - 3y + 5 = 0)

Solving
2x - y = 5
or, y = 2x - 5 - (i)

x - 3y + 5 = 0
[ Put value of y from equation (i) ]
or, x - 3(2x - 5) + 5 = 0
or, x - 6x +15 + 5 = 0
or, -5x + 20 = 0
or, 5x = 20
So, x = 4

Put value of x in equation (i), we get,
y = 2×4 - 5
So, y = 3

Hence, point of intersection of the above lines are (4,3).
So, centre (h,k) = (4,3)

When (h,k) = (4,3) and r = 5, we have,
(x - h)² + (y - k)² = r²
or, (x - 4)² + (y - 3)² = 5²
or, (x² - 8x + 16) + (y² - 6y + 9) = 25
or, x² + y² - 8x - 6y = 0 is the required equation.

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

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