The circle C has equation $$x^2 + y^2 + 4x - 2y - 11 = 0$$ Find a) the coordinates of the centre of C, b) the radius of C, c) the coordinates of the points where C crosses the y-axis, giving your answers as simplified surds. - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 2

From the standard form of the equation we found:

$(x + 2)^2 + (y - 1)^2 = 16$

The radius r is given by the square root of the right side:

r = rac{ ext{sqrt}(16)}{1} = 4

Thus, the radius of circle C is 4.

To find where the circle crosses the y-axis, we set x = 0 in the circle's equation:

$0^2 + y^2 + 4(0) - 2y - 11 = 0$

This simplifies to:

$y^2 - 2y - 11 = 0$

We can solve this using the quadratic formula:

y = rac{-b ext{±} ext{sqrt}(b^2 - 4ac)}{2a}

In this equation, a = 1, b = -2, and c = -11:

y = rac{2 ext{±} ext{sqrt}((-2)^2 - 4(1)(-11))}{2(1)}

Calculating under the square root:

$b^2 - 4ac = 4 + 44 = 48$

Thus:

y = rac{2 ext{±} ext{sqrt}(48)}{2}

This simplifies to:

$y = 1 ext{±} 2 ext{sqrt}(3)$

Therefore, the coordinates where the circle C crosses the y-axis are:

$$(0, 1 + 2 ext{sqrt}(3)) ext{ and } (0, 1 - 2 ext{sqrt}(3)).$