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 standard equation of the circle, we see that the radius is the square root of the constant term in the equation:

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

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

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

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

This simplifies to:

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

Taking the square root: $y + 7 = 7 \quad \text{or} \quad y + 7 = -7$

Thus, we have:

1. $y = 0$
2. $y = -14$

Hence, the y-coordinates where the circle C crosses the y-axis are $0$ and $-14$.

First, we need to find the gradient of the radius from the centre (1, -7) to the point (2, 0):

Gradient of the radius = $\frac{0 - (-7)}{2 - 1} = 7$

Thus, the gradient of the tangent is the negative reciprocal: Gradient of tangent = $-\frac{1}{7}$

Using point-slope form, the equation of the tangent line can be given as: $y - 0 = -\frac{1}{7}(x - 2)$

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

In the form of $ax + by + c = 0$, we can multiply through by $7$ to eliminate the fraction: $x + 7y - 2 = 0$

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