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 rearranged equation of the circle:

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

we can identify the radius.

Here, $r^2 = 16$, therefore:

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

Thus, the radius of circle C is 4.

To find the points where the circle C crosses the x-axis, 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 gives:

$x - 5 = ext{+/-}4.$

Thus, we have:

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

Therefore, the coordinates where the circle crosses the x-axis are $(9, 0)$ and $(1, 0)$.

Given that the line has a gradient of $\frac{7}{2}$ and goes through point A $(5, 0)$, we can use the point-slope form of a line equation:

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

Substituting the known values:

• $m = \frac{7}{2}$
• $(x_1, y_1) = (5, 0)$

We have:

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

Simplifying, we get:

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

Thus, the equation of the line that passes through A and touches C at point T is:

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