Only two circle theorems are examined at Core tier, and both must be quoted by name whenever they're used — a correct angle with no reason given typically loses the reasoning mark even if the number itself is right.

Any angle subtended at the circumference by a diameter is a right angle:

        C
       ╱ ╲
      ╱   ╲
     ╱     ╲
A ──────────── B   (AB is a diameter)
   ∠ACB = 90°

This holds no matter where $C$ sits on the circumference (as long as it's not at $A$ or $B$ themselves) — the angle stays exactly $90^\circ$ every time, which is what makes the rule so powerful for exam questions: spotting a diameter anywhere in a circle diagram instantly hands you a right angle to work with.

Worked example. $AB$ is a diameter of a circle centre $O$, and $C$ is on the circumference with angle $CAB = 27^\circ$. Find angle $ACB$ and angle $ABC$.

Since $AB$ is a diameter, $\angle ACB = 90^\circ$ (angle in a semicircle). Angles in triangle $ABC$ sum to $180^\circ$, so $\angle ABC = 180^\circ - 90^\circ - 27^\circ = 63^\circ$.

A tangent touching a circle at point $T$ is always perpendicular to the radius $OT$ drawn to that point of contact:

        T (point of contact)
        │
────────T──────── (tangent line)
        │ 90°
        O (centre)

Spotting this instantly turns the diagram into a right-angled triangle problem, since any line from the centre $O$ to an external point $P$, together with the radius $OT$ and the tangent $PT$, forms a right-angled triangle with the right angle sitting exactly at $T$.

Worked example. $PT$ is a tangent to a circle centre $O$, touching at $T$. $OP = 13\text{ cm}$ and radius $OT = 5\text{ cm}$. Find $PT$.

Angle $OTP = 90^\circ$, so triangle $OTP$ is right-angled at $T$ with hypotenuse $OP$. By Pythagoras:

Exam questions often place a diameter and a tangent in the same diagram, requiring both rules in sequence.

Worked example. $AB$ is a diameter of a circle centre $O$. $C$ is on the circumference, and the tangent at $C$ meets the line $AB$ extended at point $P$, outside the circle. Given $\angle BAC = 35^\circ$, find $\angle OCP$.

Since $AB$ is a diameter, $\angle ACB = 90^\circ$ (angle in a semicircle), so $\angle OCB = 180^\circ - 90^\circ - 35^\circ = 55^\circ$ using the triangle angle sum in $ABC$. The tangent at $C$ is perpendicular to the radius $OC$, so $\angle OCP = 90^\circ$ — this angle is fixed by the tangent–radius rule alone and doesn't depend on where $P$ sits along the extended diameter.