Equal chords are equidistant from the centre. If two chords in the same circle are the same length, the perpendicular (shortest) distance from the centre to each chord is identical.

The perpendicular bisector of any chord passes through the centre of the circle. This is extremely useful for locating an unknown centre: draw the perpendicular bisectors of two different chords, and their intersection point is the centre.

Drawing the two radii to the endpoints of a chord creates an isosceles triangle (both radii are equal in length). Dropping a perpendicular from the centre to the chord splits this isosceles triangle into two congruent right-angled triangles, which means it must split the chord itself into two equal halves. This is also why the perpendicular distance from the centre — used in the equal-chords fact above — is measured along this same bisecting line.

Tangents from an external point are equal in length. If two tangent lines are drawn from a single point outside the circle to touch the circle at two different points, both tangent segments have identical length — this also creates a kite shape (using the two radii and two tangents) that is symmetrical about the line joining the external point to the centre.

Questions often draw a tangent-chord diagram that needs both sets of theorems: use the symmetry facts here (equal tangents, perpendicular bisectors) to find lengths or spot isosceles triangles, then use the angle theorems from E4.7 (tangent–radius $90°$, alternate segment) to find angles within that same diagram. The tangent-radius right angle in particular is often the missing link: once you know two of the angles in the kite formed by two tangents and two radii are both $90°$, the quadrilateral angle sum ($360°$) lets you find the remaining two angles quickly.

$PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$, touching the circle at $A$ and $B$. Given $PA = 12\text{ cm}$, state $PB$ and explain your reasoning.

1. $PA$ and $PB$ are both tangents drawn from the same external point $P$.
2. Tangents from an external point are equal in length.
3. Therefore $PB = 12\text{ cm}$.

$PA$ and $PB$ are tangents from external point $P$ to a circle with centre $O$, touching at $A$ and $B$. Angle $APB = 46°$. Find angle $AOB$.

1. $OA \perp PA$ and $OB \perp PB$ (tangent–radius theorem), so angle $OAP =$ angle $OBP = 90°$.
2. The quadrilateral $OAPB$'s interior angles sum to $360°$.
3. $\text{angle } AOB = 360° - 90° - 90° - 46° = 134°$.