Question: Find the equation of a circle concentric with the circle x² + y² -6x +y = 1 and passing through the point (4,-2).

Solution:

Since two circles are concentric, the coordinates of the centre of the circle is same.

We know,
Equation of one of the concentric circle is x² + y² -6x + y -1 = 0.
Comparing above equation with x² + y² + 2gx + 2fy + c = 0, we get,

-g = 3, -f = -1/2

So, centre (h,k) = (-g,-f) = (3,-1/2)

When (h,k) = (3, -1/2) and passing point (x,y) = (4,-2), we get,
(x - h)² + (y - k)² = r²
or, (4 - 3)² + {-2 - (-1/2)}² = r²
or, 1 + 9/4 = r²
So, r² = 13/4

Now,
Square of the radius of the required circle (r²) = 13/4 unit²
Coordinates of the centre of the circle (h,k) = (3, -1/2)

Equation of the required circle is given by,
(x - h)² + (y - k)² = r²
or, (x -3)² + {y - (-1/2)}² = 13/4
or, x² -6x + 9 + (y +1/2)² = 13/4
or, x² -6x + 9 + y² + y + 1/4 = 13/4
or, (4x² - 24x + 36 + 4y² + 4y +1)/4 = 13/4
or, 4x² + 4y² -24x + 4y + 24 = 0
or, x² + y² - 6x + y + 6 = 0 is the required equation.

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

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