A circle with centre C has equation x^2 + y^2 + 2x + 6y - 40 = 0 . Express this equation in the form (x - a)^2 + (x - b)^2 = r^2. Find the co-ordinates of C and the radius of the circle.

We can express the equation of any circle in the form (x - a)^2 + (y- b)^2 = r^2. The centre of the circle is then the point (a, b) and its radius is r. 

If we multiply out the brackets in the equation above, we get

x^2 + y^2 - 2ax - 2by + a^2 + b^2 = r^2.

Bringing the r^2 to the left side by subtracting it from both sides, we have

x^2 + y^2 - 2ax - 2by + a^2 + b^2 - r^2 = 0.    (1)

Let's compare this to the equation we are given,

x^2 + y^2 + 2x + 6y - 40 = 0.                            (2)

We compare the coefficients of the x's and y's in each equation to find a and b. In equation (2) we have 2x and we have 6y, whilst in (1) we have -2ax and -2by. So we must have 2x = -2ax and 6y = -2by. Therefore a = -1 and b = -3.

So a^2 = 1 and b^2 = 9.

Substituting these values of a and b into (1) gives us 

x^2 + y^2 + 2x + 6y + 1 + 9 - r^2 

= x^2 + y^2 + 2x + 6y + 10 - r^2 

= 0.

This looks almost like (2). The final step is making 10 - r^2 = -40. This gives us r^2 = 50, so r = (50)^(1/2).

And now we substitute our values of a, b and r^2 into (x - a)^2 + (y - b)^2 = r^2 to get the equation:

(x + 1)^2 + (y + 3)^2 = 50.

So C, the centre of the circle, is (-1, -3), and the radius of the circle is (50)^(1/2).