The circle C has equation $$x^2 + y^2 - 2x + 14y = 0$$ Find a) the coordinates of the centre of C, b) the exact value of the radius of C, c) the y coordinates of the points where the circle C crosses the y-axis - Edexcel - A-Level Maths Pure - Question 7 - 2018 - Paper 4

From the equation we derived above:

$(x - 1)^2 + (y + 7)^2 = 50$

The radius r is the square root of 50:

$r = ext{sqrt}{50} = 5 ext{sqrt}{2}$

Thus, the exact value of the radius is $5 ext{sqrt}{2}$.

To find where the circle crosses the y-axis, we set $x = 0$ and substitute into the equation:

$(0 - 1)^2 + (y + 7)^2 = 50$

This simplifies to:

$1 + (y + 7)^2 = 50$ $(y + 7)^2 = 49$

Taking the square root gives:

$y + 7 = ext{±7}$

Thus, we find:

$y = 0 ext{ or } y = -14$

The y-coordinates where the circle crosses the y-axis are 0 and -14.

To find the tangent line at the point (2, 0), we first find the gradient of the radius at this point. The centre of the circle is (1, -7), so the slope of the radius is:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-7)}{2 - 1} = 7$

The tangent line is perpendicular to the radius, so its slope is:

$m_{tangent} = -\frac{1}{7}$

Using the point-slope form of the line equation:

$y - y_1 = m(x - x_1)$

We substitute (2, 0):

$y - 0 = -\frac{1}{7}(x - 2)$

Rearranging gives:

$y = -\frac{1}{7}x + \frac{2}{7}$

To write this in the form $ax + by + c = 0$, we can rearrange:

$\frac{1}{7}x + y - \frac{2}{7} = 0$

Multiplying through by 7 to eliminate the fraction yields:

$x + 7y - 2 = 0$

Thus, the equation of the tangent line is $x + 7y - 2 = 0$.