Question 45348
The standard equation of a circle has the form 

(x - h)^2 + (y - k)^2 = r^2 
where h and k are the x- and y-coordinates of the center of the circle and r is the radius. 

Center of the circle is at (-1, 5)

Hence substituting in the equation of circle:
(x - (-1))^2 + (y - 5)^2 = r^2
x^2 + 2x + 1 + y^2 - 5y + 25 = r^2

Now it is given that (-4, -6) is a point on the circle, hence it will satisfy the equation of the circle

Substiute x = -4 and y = -6 in the above equation:
(-4)^2 + 2(-4) + 1 + (-6)^2 - 5(-6) + 25 = r^2
r^2 = 16 - 8 + 1 + 36 + 30 + 25
r^2 = 100

Substituting this value of r^2 in the equation for the circle:
x^2 + 2x + y^2 - 5y + 26 = 100
x^2 + 2x + y^2 - 5y = 74

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