Let ABCD be a rectangle with AB = 72 cm and BC = 30 cm. A circle passing through points A and C cuts the side AB at P such that AP = 56 cm. The radius, in cm, of the circle is

Coordinate setup: $$A = (0,0)$$, $$B = (72,0)$$, $$C = (72,30)$$, $$D = (0,30)$$, $$P = (56, 0)$$.

Circle through $$A, P, C$$: $$x^2 + y^2 + Dx + Ey + F = 0$$.

• $$A$$: $$F = 0$$.
• $$P$$: $$56^2 + 56D = 0 \Rightarrow D = -56$$.
• $$C$$: $$72^2 + 30^2 - 56 \cdot 72 + 30E = 0 \Rightarrow 30E = -2052 \Rightarrow E = -68.4$$.

Radius$$^2 = (D/2)^2 + (E/2)^2 - F = 28^2 + 34.2^2 = 784 + \tfrac{29241}{25} = \tfrac{48841}{25}$$, so radius $$= \dfrac{221}{5}$$.

Geometric alternative: the centre lies on the perpendicular bisectors of chords $$AP$$ and $$AC$$. Midpoint of $$AP$$ is $$(28, 0)$$ and $$AP$$ is horizontal, so the perpendicular bisector is $$x = 28$$. Midpoint of $$AC$$ is $$(36, 15)$$ with slope $$\tfrac{5}{12}$$, so its perpendicular bisector has slope $$-\tfrac{12}{5}$$ through $$(36, 15)$$. Setting $$x = 28$$: $$y = 15 + \tfrac{12}{5}(8) = \tfrac{171}{5}$$. So centre $$= (28, \tfrac{171}{5})$$, giving the same radius $$\tfrac{221}{5}$$.