Find the coordinates of the centre of the circle with equation: x^2 + y^2 − 2*x + 14*y = 0

method: complete the squares for (x^2 - 2x ) and ( y^2 + 14y ) using the formula (b/2)^2 (where the general equation is ax^2 + bx) so that the equation becomes ax^2 + bx + (b/2)^2first, rearrange the original equation: (x^2 - 2x ) + (y^2 + 14y) = 0then use the completing the squares method on each bracketed section: (x^2 - 2x + 1) + (y^2 + 14y + 49) = 0factorise each bracketed section: (x - 1)^2 + (y + 7)^2 = 0centre of the circle (a,b) is found when the circle equation is in this format: (x - a)^2 + (y - b)^2 = r^2 (where r is the radius of the circle). Using this, the centre of the circle is (1,-7)Answer: centre of the circle is (1, -7)