The point A has co-ordinates (8, 6) and O is the origin - Leaving Cert Mathematics - Question Question 1 - 2013

The center of circle c₂ can be found by averaging the coordinates of points O (0, 0) and A (8, 6):

$ext{Center} = \left( \frac{8 + 0}{2}, \frac{6 + 0}{2} \right) = (4, 3).$

Next, the diameter of c₂ is equal to the distance |OA|, which we found earlier to be 10. Therefore, the radius r of c₂ is:

$r = \frac{10}{2} = 5.$
Using the equation of a circle:

$(x - h)^2 + (y - k)^2 = r^2$
we substitute (h, k) with (4, 3) and r with 5:

$(x - 4)^2 + (y - 3)^2 = 5^2$ $= 25.$
Thus, the equation of c₂ is:

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

To find where the circle c₂ intersects the x-axis, we set y = 0 in the equation of c₂:

$(x - 4)^2 + (0 - 3)^2 = 25$ $(x - 4)^2 + 9 = 25$ $(x - 4)^2 = 25 - 9$ $(x - 4)^2 = 16$ Taking the square root of both sides gives:

$x - 4 = 4 ext{ or } x - 4 = -4$ Thus,

$x = 8 ext{ or } x = 0.$
The corresponding coordinates at these points are:

1. For x = 8, P(8, 0).
2. For x = 0, P(0, 0).

The co-ordinates of P are:

Answer:
P(8, 0) and P(0, 0)