What is the angle between a tangent and the radius at the point of contact?

90 degrees. A tangent is perpendicular to the radius drawn to the point where it touches the circle, so the angle there is always a right angle. That right angle is what turns the picture into a right-angled triangle, so Pythagoras or an angle-chase can start. Treating the tangent and radius as ordinary crossing lines is the trap.

PQ is a diameter and QR is a tangent at Q with PQ equal to QR and PR equal to 10. What is the radius?

The radius is (5 root 2) over 2, about 3.54 cm. The tangent QR meets the diameter PQ at 90 degrees, so triangle PQR is right-angled with hypotenuse PR = 10. Since PQ = QR, Pythagoras gives 2 times PQ squared = 100, so PQ = root 50 = 5 root 2, which is the diameter. Halving gives the radius (5 root 2) over 2, about 3.54. The trap is to stop at 5 root 2 (about 7.07), the diameter, instead of halving.

Two tangents are drawn from a point P with a 24 degree angle between them. What is the angle at the centre?

156 degrees. Each tangent meets its radius at the point of contact at 90 degrees, so the quadrilateral formed by the two radii, the two tangent segments and the centre has two right angles. The angles in a quadrilateral sum to 360, so the centre angle is 360 minus 90 minus 90 minus 24 = 156 degrees. Missing one of the two right angles is the trap.

GCSE Maths Higher · AQA · Circle theorems

Tangent perpendicular to radius: plant the right angle first

The circle-theorem fact that unlocks the whole question: a tangent meets the radius drawn to the point of contact at exactly 90 degrees. That single right angle is what turns a circle diagram into a right-angled triangle, so Pythagoras or an angle-chase can begin. Without it, the question simply stalls.

The trap is to treat the tangent and the radius as ordinary crossing lines and never mark the $90^\circ$ . So a Pythagoras set-up never forms, a kite angle-chase never closes, and a multi-mark “show that” proof never starts. The fix is one habit: wherever a tangent touches the circle, draw the radius to that point and plant a right angle there, every time.

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How to spot it in your own work

You drew a tangent and the radius to the contact point but did not mark the $90^\circ$ between them, so no right-angled triangle formed.
You said a tangent problem “cannot be done” or “has no right angle given,” when the right angle is the tangent-radius fact you are expected to supply.
You found the diameter $PQ = 5\sqrt{2}$ but called it the radius, forgetting to halve it to $\tfrac{5\sqrt{2}}{2}\approx 3.54$ .
With two tangents from one point you used only one right angle (or none), so the centre angle came out as $204^\circ$ instead of $156^\circ$ .

An exam question that triggers it

Here is NOV23 Paper 3 Question 16, reproduced in structure:

PQ is a diameter of a circle.

QR is a tangent to the circle at Q.

$PQ = QR$ and $PR = 10\text{ cm}$ .

Work out the radius of the circle.

The misconception leaves the problem stuck, because no right angle is placed. The fix: the tangent QR meets the diameter PQ at $90^\circ$ , so triangle PQR is right-angled with hypotenuse $PR = 10$ . Then $2\,PQ^2 = 100$ , giving $PQ = \sqrt{50} = 5\sqrt{2}$ (the diameter) and a radius of $\tfrac{5\sqrt{2}}{2}\approx 3.54\text{ cm}$ .

Why students fall for this

The right angle is the one piece of information the diagram does not draw. A tangent and a radius cross at the contact point, but the $90^\circ$ is a fact you have to recall and add yourself. Students who wait for the picture to show it never place it, so the triangle never becomes right-angled.

It also hides behind the word “tangent.” A diameter ending at the contact point, or a chord through it, looks like just another line. Recognising that the diameter lies along the radius, so the tangent is perpendicular to it too, is the step that gets skipped.

Finally, the right angle gets put in the wrong place. A chord meeting a radius has no special angle, and two radii to a chord make an isosceles triangle, not a right angle. Reaching for $90^\circ$ where it does not belong, or missing one of the two right angles when two tangents meet at a point, is the same error wearing a different coat.

The fix: Find the tangent, plant the right angle, then chase

Step 1: find the tangent contact point. Locate where a tangent touches the circle and draw the radius to that exact point.

Step 2: plant the right angle. Mark $90^\circ$ between the tangent and that radius. A diameter ending at the contact point counts, because it lies along the radius.

Step 3: use it. The right angle makes a right-angled triangle, so Pythagoras runs: $2\,PQ^2 = 100 \Rightarrow PQ = 5\sqrt{2}$ , then halve for the radius, $\tfrac{5\sqrt{2}}{2}\approx 3.54$ .

Step 4: count all the right angles. Two tangents from one point give two right angles, so the quadrilateral angle-chase gives $360 - 90 - 90 - 24 = 156^\circ$ at the centre. A chord or a radius pair gives none, so do not invent one.

Worked example

NOV23·3·16 structure: PQ is a diameter, QR a tangent at Q, $PQ = QR$ , $PR = 10$ . Find the radius.

Right angle: the tangent QR meets the diameter PQ at the contact point, so $\angle PQR = 90^\circ$ .
Pythagoras, equal legs.
$PQ^2 + QR^2 = PR^2 \Rightarrow 2\,PQ^2 = 100 \Rightarrow PQ = \sqrt{50} = 5\sqrt{2}$
Halve the diameter. $r = \tfrac{5\sqrt{2}}{2}\approx 3.54\text{ cm}$ . Trap: stopping at $5\sqrt{2}\approx 7.07$ , the diameter.

JUN24·1·22 structure: AP and BP are tangents from an external point P, with $\angle APB = 24^\circ$ . Find $\angle AOB$ at the centre, then the angle $x$ at the circumference.

Two right angles. $\angle OAP = \angle OBP = 90^\circ$ , one at each contact point.
Quadrilateral chase. $\angle AOB = 360 - 90 - 90 - 24 = 156^\circ$ .
Halve for the circumference. $x = \tfrac{156}{2} = 78^\circ$ . Trap: using only one right angle gives $204^\circ$ .

JUN22·3·18, a 5-mark show that: DC is a tangent at C, the angle ACD between chord CA and the tangent is $83^\circ$ , and $\angle BAC = 28^\circ$ . Show $\angle ABO : \angle ACO = 3 : 1$ .

Tangent perpendicular to radius. $\angle OCD = 90^\circ$ , so $\angle ACO = 90 - 83 = 7^\circ$ .
Isosceles radii. $OA = OC$ , so $\angle OAC = 7^\circ$ ; then $\angle OAB = 28 - 7 = 21^\circ$ , and $OA = OB$ gives $\angle ABO = 21^\circ$ .
Conclude. $\angle ABO : \angle ACO = 21 : 7 = 3 : 1$ . Trap: no right angle means $\angle ACO$ is never found.

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Common questions

What angle does a tangent make with the radius at the point of contact?: $90^\circ$ . A tangent is perpendicular to the radius drawn to the point where it touches the circle. That right angle is what makes a right-angled triangle, so Pythagoras or an angle-chase can start. Treating the tangent and radius as ordinary crossing lines is the trap.
PQ is a diameter and QR a tangent at Q, with $PQ = QR$ and $PR = 10$ . What is the radius?: $\tfrac{5\sqrt{2}}{2}\approx 3.54\text{ cm}$ . The tangent meets the diameter at $90^\circ$ , so $2\,PQ^2 = 100$ and $PQ = \sqrt{50} = 5\sqrt{2}$ , the diameter. Halve it for the radius. Stopping at $5\sqrt{2}\approx 7.07$ is the trap.
Two tangents from a point have $24^\circ$ between them. What is the centre angle?: $156^\circ$ . Each tangent meets its radius at $90^\circ$ , so the quadrilateral has two right angles and $\angle AOB = 360 - 90 - 90 - 24 = 156^\circ$ . Missing one right angle gives $204^\circ$ .

Related misconceptions

Angle at the centre not doubledThe angle at the centre is twice the angle at the circumference on the same arc. Treating them as equal is a top-grade trap.
Cyclic quadrilateral and alternate segmentOpposite angles of a cyclic quadrilateral sum to 180, and the alternate segment angle needs the right sides checked.

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