Exam practice

Half the chord: $AM = \sqrt{5^2 - 4^2} = \sqrt{9} = 3$ cm (the perpendicular OM bisects the chord). The chord is 6 cm long.

The radius is 17 cm. Half the chord: $\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$ cm. The chord is 30 cm long.

Half the chord is 10 cm. Distance: $\sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24$ cm. The chord is 24 cm from the center.

Half the chord is 24 cm. Distance: $\sqrt{26^2 - 24^2} = \sqrt{676 - 576} = \sqrt{100} = 10$ cm. The chord is 10 cm from the center.

Half the chord is 12 cm, so the radius is $\sqrt{10^2 + 12^2} = \sqrt{244} = 2\sqrt{61} \approx 15.62$ cm. The diameter is $4\sqrt{61} \approx 31.24$ cm.

Let CD, the longer chord, be x cm from the center. Then AB, the shorter, is farther away: (x + 3) cm. Write $r^2$ twice.

Let ON = x cm (to CD); then OM = (x + 3) cm (to AB). Half of CD is 6 cm and half of AB is 3 cm, and both right-angled triangles have the radius as hypotenuse:
$$\begin{aligned} x^2 + 6^2 &= (x + 3)^2 + 3^2 \\ x^2 + 36 &= x^2 + 6x + 9 + 9 \\ 18 &= 6x \\ x &= 3 \end{aligned}$$
$r^2 = 3^2 + 6^2 = 45$, so $r = \sqrt{45} = 3\sqrt5 \approx 6.71$ cm.

Distance of PQ: $\sqrt{5^2 - 4^2} = 3$ cm. Distance of XY: $\sqrt{5^2 - 3^2} = 4$ cm. They are on opposite sides, so they are 3 + 4 = 7 cm apart.

No. In a circle of radius 2 cm the longest chord is the diameter, 4 cm. A and B are 5 cm apart, so they can’t both lie on such a circle. (Even the circle with its center at M, the middle of AB, falls short of both by 0.5 cm, and any other center is farther from A or B. The smallest circle through A and B has radius 2.5 cm.)

Proved.

Draw the perpendicular bisector of AB with a compass (arcs of the same radius, more than half of AB, from A and from B, crossing on both sides; join the crossings), and in the same way the perpendicular bisector of BC. The perpendicular bisector of every chord passes through the center, so the point O where the two cross is the center. Check: OA, OB and OC measure the same, the radius.