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 Mechanics - Question 7 - 2021 - Paper 2

From the rearranged equation:

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

The radius $r$ can be derived from the right-hand side as:

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

Thus, in terms of $p$, the radius of the circle is:

r = rac{ ext{sqrts}+25 + p}{25}.

To determine two possible values of $p$, we need to consider the conditions under which the circle intersects the coordinate axes at three points. This occurs when the circle is tangent to one of the axes.

1. When the circle passes through the origin, we set: $r^2 = 25 + p$ This gives us: $0 = 25 + p$ Thus, $p = -25$.

2. For the case where the circle touches the x-axis, we substitute $y=0$ into the equation: $(x-3)^2 + (0-4)^2 = r^2$ The equation becomes: $(x-3)^2 + 16 = 25 + p$ Solving for $p$: $16 = p$ Thus, $p = -9$.

The two possible values of $p$ are therefore: $p = -25$ and $p = -9$.