The circle C, with centre at the point A, has equation $x^2 + y^2 - 10x + 9 = 0.$ Find (a) the coordinates of A, (b) the radius of C, (c) the coordinates of the points at which C crosses the x-axis - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 2

From the standard form of the circle equation,

$(x - 5)^2 + y^2 = 16$

we note that the radius $r$ is given by the square root of the constant on the right side:

$r = ext{sqrt}(16) = 4$

Radius of C: 4

To find the x-intercepts, we set $y = 0$ in the equation of the circle:

$(x - 5)^2 + 0^2 = 16$

This simplifies to:

$(x - 5)^2 = 16$

Taking the square root of both sides gives:

x - 5 = m{4} ext{ or } x - 5 = -4

Solving these equations:

1. $x - 5 = 4 \rightarrow x = 9$
2. $x - 5 = -4 \rightarrow x = 1$

Thus, the coordinates of the points where C crosses the x-axis are:

Coordinates: (9, 0) and (1, 0)

Since we know the gradient of the tangent line at point T is $\frac{7}{2}$, we can use point A(5, 0) to find the equation of the line. The point-slope formula is:

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

Substituting $(x_1, y_1) = (5, 0)$ and $m = \frac{7}{2}$ yields:

$y - 0 = \frac{7}{2}(x - 5)$

Simplifying, we arrive at:

$y = \frac{7}{2}x - \frac{35}{2}$

Therefore, the equation of the line is:

Equation of the line: $y = \frac{7}{2}x - \frac{35}{2}$