A circle has equation $$x^2 + y^2 - 6x - 8y = p$$ 7 (a) (i) State the coordinates of the centre of the circle - AQA - A-Level Maths Pure - Question 7 - 2021 - Paper 2

From the standard form of the circle's equation, we have:

$(x - 3)^2 + (y - 4)^2 = p + 25$.

The radius $r$ is given by the expression under the square root as:

$r = ext{sqrt}(p + 25)$.

Therefore, the radius of the circle in terms of $p$ is: r = rac{ ext{sqrt}{25 + p}}{1}.

For the circle to intersect the coordinate axes at exactly three points, one of the axes must be tangent to the circle, and the other must intersect it.

1. If the circle passes through the origin $(0, 0)$, substituting into the circle's equation: $0^2 + 0^2 - 6(0) - 8(0) = p$ $p = 0$.

2. For the circle to just touch the x-axis (i.e., the circle's edge is tangent to the x-axis), we set $y=0$ in the standard form: $(x - 3)^2 + (0 - 4)^2 = p + 25$, $(x - 3)^2 + 16 = p + 25$. Thus, solving: $(x - 3)^2 = p + 9$

3. Setting $p + 9 = 0$ gives: $p = -9$.

In conclusion, the two possible values of $p$ are $0$ and $-9$.