Question: Find the equation of the circle whose centre is (7,-5) and it passes through the centre of the circle 2x² + 2y² +4x +6y -14 = 0.

Solution:

Given,
Co-ordinates of the centre of the required circle (h,k) = (7,-5)

Equation of the other circle is 2x² + 2y² + 4x + 6y - 14 = 0
or, 2 (x² + y² + 2x + 3y - 7) = 0
or, x² + y² + 2x + 3y -7 = 0 -(i)

Comparing equation (i) with the general equation of circle x² + y² + 2gx + 2fy + c = 0, we get,
-g = -1, -f = - 3/2
So, centre of the other circle is (-1, -3/2)

According to the question,
(-1,-3/2) is the passing point of the required circle.
So, (-1, -3/2) = (x,y)

When (x,y) = (-1, -3/2) and (h,k) = (7, -5), using formula for circle equation, we get,
(x - h)² + (y - k)² = r²
or, (-1 - 7)² + {-3/2 - (-5)}² = r²
or, 64 + 49/4 = r²
or, r² = 305/4

And,
When (h,k) = (7,-5) and r² = 305/4, equation of circle is given by,
(x - h)² + (y - k)² = r²
or, (x -7)² + {y - (-5)}² = 305/4
or, (x -7)² + (y +5)² = 305/4
or, x² - 14x + 49 + y² + 10y + 25 = 305/4
or, x² + y² - 14x + 10y + 74 = 305/4
or, 4(x² + y² - 14x + 10y + 74) = 305
or, 4x² + 4y² - 56x + 40y + 296 = 305
So, 4x² + 4y² - 56x + 40y - 9 = 0 is the required equation.

Hence, the required equation of the mentioned circle is 4x² + 4y² - 56x + 40y - 9 = 0.

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