The diagram shows two points \(A\) and \(B\) with \(AB = 10\) cm. The locus of points 6 cm from \(A\) meets the locus of points 8 cm from \(B\) at the point...

The locus of points a fixed distance from a point is a circle centred on that point, so \(P\) lies 6 cm from \(A\) and 8 cm from \(B\), while \(AB = 10\) cm. That gives a triangle with all three sides known.

(a) The converse of Pythagoras states that if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled, with the right angle opposite the longest side. Test the two shorter sides:

\(6^2 + 8^2 = 36 + 64 = 100\) [M1]

\(100 = 10^2 = AB^2\), so by the converse of Pythagoras angle \(APB\) is a right angle [A1]

The right angle is at \(P\) because \(AB\) is the longest side and the angle at \(P\) is the one facing it.

(b) Every point that sees the segment \(AB\) at a right angle lies on the same circle:

A circle with \(AB\) as its diameter [B1] oe

This is the circle theorem that the angle in a semicircle is \(90^\circ\), read in reverse. Since \(AB\) is 10 cm, the circle has radius 5 cm and is centred on the midpoint of \(AB\). Note that this must be quoted as a circle on \(AB\) as diameter; a circle of radius 10 cm, or one centred at \(A\), does not have the required property.