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Year 10 Maths Advanced Circle Geometry

Tangent Properties

20 practice questions 2 video lessons Theory + worked examples

A tangent touches a circle at exactly one point and is perpendicular to the radius there. Two tangents drawn from an outside point are equal in length, and the angle between a tangent and a chord equals the angle in the alternate segment. In NSW Year 10 Advanced maths these properties are used to find angles and lengths and in proofs (NESA outcome MA5‑CIR‑P‑01).

You will find angles using a tangent and a radius, the alternate segment theorem and equal tangents, and find lengths when a circle sits inside a triangle or a quadrilateral. You will show that the sums of opposite sides of a quadrilateral drawn around a circle are equal and find the exact length of tangents from a point. Harder questions prove that tangents from a point are equal and solve problems with two circles that touch. It is part of Year 10 Advanced Circle Geometry.

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  • Tangent Properties - Video - Example with tangent and radius | Circles Watch
  • Tangent Properties - Video - All GCSE circle theorems & proofs Watch
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Theory

A tangent touches a circle at exactly one point and is perpendicular to the radius there. Two tangents from an external point are equal, and the angle between a tangent and a chord equals the angle in the alternate segment. This Year 10 Advanced maths page uses these facts to find angles and lengths and to write proofs.

A tangent is a line that touches a circle at exactly one point, the point of contact. At any point on a circle there is exactly one tangent, and it is perpendicular to the radius drawn to that point.

From a point \(P\) outside a circle you can draw two tangents. The tangent lengths \(PA\) and \(PB\), from \(P\) to the points of contact, are equal.

A chord drawn from the point of contact splits the circle into two segments. The alternate segment is the one on the other side of the chord from the angle you are looking at.

A circle drawn inside a triangle or quadrilateral so that it touches every side is inscribed; each side is a tangent. Two circles touch if they meet at one point and share a tangent there.

It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.

Two tangents from an external pointA circle with centre O and an external point P. Tangents PA and PB touch the circle at A and B and are marked equal. Radii OA and OB meet the tangents at right angles. The dashed line OP bisects the angle APB. O A B P
Tangent \(\perp\) radius at the point of contact. The two tangents from \(P\) are equal and \(OP\) bisects \(\angle APB\).
Alternate segment theoremA tangent SAT touches a circle at A. Chords AB and AC are drawn and B and C are joined. The angle between the tangent AT and chord AC is theta and equals the angle ABC in the alternate segment. The angle between AS and AB is phi and equals the angle ACB. θ θ φ φ A B C S T
Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.

For tangents \(PA\) and \(PB\) from \(P\) to a circle with centre \(O\):

\[OA \perp PA \qquad PA = PB \qquad \angle APB + \angle AOB = 180^\circ\]
PA=PB,∠APB+∠AOB=180

The last result comes from the angle sum of quadrilateral \(OAPB\), which has two right angles. For a secant from \(P\) cutting the circle at \(A\) and \(B\), and a tangent \(PT\):

\[PT^2 = PA \times PB\]
PT2=PA×PB
Alternate segment theorem. The angle between a tangent and a chord drawn from the point of contact equals the angle at the circumference in the alternate segment.
Inscribed circles. In a triangle, call the three tangent lengths \(p\), \(q\), \(r\); each side is the sum of two of them, so \(p + q + r\) is half the perimeter. In a quadrilateral whose four sides all touch a circle, the opposite sides have equal sums: \(AB + CD = BC + DA\).
Exact values. \(OP\) bisects \(\angle APB\). If \(\angle APB = 90^\circ\), then \(\triangle OAP\) has a \(45^\circ\) angle at \(P\) and \(PA = OA\). With \(30^\circ\), \(45^\circ\) or \(60^\circ\) at \(P\), use the exact trigonometric ratios to give exact (surd) lengths. The area of kite \(OAPB\) is twice the area of \(\triangle OAP\).

How to use tangent properties

  1. Mark a right angle wherever a radius meets a tangent, and equal ticks on the two tangents from each external point.
  2. Look for the shape: quadrilateral \(OAPB\) (angles add to \(360^\circ\)), isosceles triangle \(PAB\), right-angled triangle \(OAP\), or a chord from the point of contact (alternate segment).
  3. Calculate with a reason at every step.

Problem types

  1. Chains of tangents: when several tangents meet a circle, each external point gives a pair of equal lengths. Split each side into its two tangent lengths and add or subtract.
  2. Inscribed circle: write each side as a sum of two unknown tangent lengths, add the three equations, then subtract (Example 2).
  3. Lengths: use Pythagoras' theorem in the right-angled triangle radius-tangent-\(OP\). If the distance from \(P\) to the circle is given, \(OP = r + \text{distance}\) (Example 4). If an angle is given, use trigonometry.
  4. Tangent and secant from one point: \(PT^2 = PA \times PB\), where \(PB\) is the whole secant. If the parts are given in terms of \(k\), form an equation and keep the positive solution.
  5. Algebra with angles: set two expressions equal using a theorem (e.g. alternate segment), or use the angle sum of \(OAPB\).
  6. Proofs: join the centre to the points of contact and to \(P\). Right angles plus equal radii and a common hypotenuse suggest RHS congruence. For parallel lines, use the alternate segment theorem to find a pair of equal alternate angles. For two circles that touch, the common tangent at the point of contact gives equal tangents from one point to both circles.
Example 1 — Two tangents and the centre
\(PA\) and \(PB\) are tangents to a circle with centre \(O\), and \(\angle APB = 54^\circ\). \(C\) lies on the major arc \(AB\). Find
(i) \(\angle AOB\)
(ii) \(\angle ACB\)
(iii) \(\angle PAB\).
Solution
Tangents PA and PB with angle APB of 54 degreesA circle with centre O. Tangents from P touch the circle at A and B, and angle APB is 54 degrees. Radii OA and OB meet the tangents at right angles. C is on the major arc, and angles AOB and ACB are unknown. 54° ? ? O A B C P

(i) A tangent is perpendicular to the radius, so \(\angle OAP = \angle OBP = 90^\circ\). Angle sum of quadrilateral \(OAPB\):

\(\angle AOB\)\(=\)\(360^\circ - 90^\circ - 90^\circ - 54^\circ\)
\(\)\(=\)\(\mathbf{126^\circ}\)
∠AOB=360−234=126

(ii) Angle at the circumference is half the angle at the centre: \(\angle ACB = \mathbf{63^\circ}\).

(iii) \(PA = PB\) (tangents from an external point), so \(\triangle PAB\) is isosceles:

\(\angle PAB\)\(=\)\(\dfrac{180^\circ - 54^\circ}{2} = \mathbf{63^\circ}\)

\(\angle PAB = \angle ACB\), exactly as the alternate segment theorem says.

Example 2 — Equal tangents in a triangle
A circle is drawn inside \(\triangle ABC\), touching \(BC\) at \(D\), \(CA\) at \(E\) and \(AB\) at \(F\). \(AB = 11\) cm, \(BC = 14\) cm and \(CA = 9\) cm. Find \(AF\), \(BD\) and \(CE\).
Solution
Circle inscribed in a triangleTriangle ABC with BC 14 cm, AB 11 cm and CA 9 cm. A circle inside touches BC at D, CA at E and AB at F. A B C D E F 14 cm 11 cm 9 cm

Tangents from an external point are equal: let \(AF = AE = p\), \(BF = BD = q\) and \(CD = CE = r\).

\(p + q\)\(=\)\(11\)
\(q + r\)\(=\)\(14\)
\(p + r\)\(=\)\(9\)

Add all three: \(2(p + q + r) = 34\), so \(p + q + r = 17\). Subtract each equation from this:

\(p\)\(=\)\(17 - 14 = 3\)
\(q\)\(=\)\(17 - 9 = 8\)
\(r\)\(=\)\(17 - 11 = 6\)
p=3,q=8,r=6

\(AF = \mathbf{3}\) cm, \(BD = \mathbf{8}\) cm, \(CE = \mathbf{6}\) cm. Check: \(8 + 6 = 14 = BC\).

Example 3 — Alternate segment theorem
\(SAT\) is a tangent to a circle with centre \(O\) at \(A\). \(B\) and \(C\) are on the circle with \(\angle TAC = 64^\circ\) and \(\angle SAB = 50^\circ\). Find \(\angle ABC\), \(\angle ACB\) and \(\angle BOC\), giving reasons.
Solution
Tangent SAT with chords AB and ACA circle with centre O touches the tangent SAT at A. The angle between AT and chord AC is 64 degrees and the angle between AS and chord AB is 50 degrees. Angles ABC, ACB and BOC are unknown. 64° 50° O A B C S T

The angle between a tangent and a chord equals the angle in the alternate segment:

\(\angle ABC\)\(=\)\(\angle TAC = \mathbf{64^\circ}\)
\(\angle ACB\)\(=\)\(\angle SAB = \mathbf{50^\circ}\)

Angles on the straight line \(SAT\):

\(\angle BAC\)\(=\)\(180^\circ - 64^\circ - 50^\circ = 66^\circ\)

Angle at the centre is twice the angle at the circumference on arc \(BC\):

\(\angle BOC\)\(=\)\(2 \times 66^\circ = \mathbf{132^\circ}\)
∠BOC=2×66=132
Example 4 — Find the radius
The point \(P\) is outside a circle with centre \(O\) and radius \(r\) cm. The nearest point of the circle to \(P\) is \(Q\), with \(PQ = 10\) cm, and the tangent from \(P\) touches the circle at \(T\) with \(PT = 20\) cm. Find \(r\).
Solution
Tangent PT of 20 cm and nearest distance PQ of 10 cmPart of a circle with centre O and radius r. P is outside the circle on the line through O. Q is the nearest point of the circle to P, with PQ 10 cm. The tangent PT is 20 cm and meets the radius OT at right angles. O T Q P r 20 10

\(Q\) lies on \(OP\), so \(OP = r + 10\). The tangent is perpendicular to the radius, so \(\angle OTP = 90^\circ\). Pythagoras' theorem in \(\triangle OTP\):

\((r + 10)^2\)\(=\)\(r^2 + 20^2\)
\(r^2 + 20r + 100\)\(=\)\(r^2 + 400\)
\(20r\)\(=\)\(300\)
\(r\)\(=\)\(\mathbf{15}\)
(r+10)2=r2+202

Check: \(15^2 + 20^2 = 625 = 25^2\), and \(OP = 15 + 10 = 25\).

Common pitfalls

The right angle is between the tangent and the radius. It is at the point of contact, not at the external point \(P\) and not between the tangent and a chord.
Use the alternate segment, not the same one. The angle between tangent \(AT\) and chord \(AC\) equals the angle at \(B\) on the other side of \(AC\). Matching it to an angle on the same side as the tangent angle is a common slip.
Tangent lengths go to the point of contact. \(PA = PB\) measures from \(P\) to where each tangent touches, not to where a line crosses the circle a second time.
Distance to the circle is not the distance to the centre. If \(P\) is \(10\) cm from the nearest point of the circle, then \(OP = r + 10\), not \(10\).

Frequently asked questions

Why is a tangent perpendicular to the radius?

The radius to the point of contact is the shortest distance from the centre to the tangent line, because every other point on the tangent is outside the circle. The shortest distance from a point to a line is the perpendicular.

Are two tangents from the same point equal?

Yes. Tangents drawn from an external point to the points of contact are equal in length. Joining the centre to the two points of contact and to the external point gives two congruent right-angled triangles (RHS).

What is the alternate segment theorem?

The angle between a tangent and a chord drawn from the point of contact equals the angle at the circumference in the alternate segment, the segment on the other side of the chord.

How do you find the radius from a tangent?

Draw the radius to the point of contact, which makes a right angle with the tangent. Then use Pythagoras' theorem in the triangle formed by the radius, the tangent and the line from the external point to the centre.

What is the tangent-secant theorem?

If a tangent \(PT\) and a secant through \(A\) and \(B\) are drawn from the same external point \(P\), then \(PT^2 = PA \times PB\).

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