Tangent Properties of a Circle

Worked example

1. 1

   $\angle OBA=24^{\circ }$ — Given. O is the centre and TA is the tangent at A.

2. 2

   $\angle OAB=24^{\circ }$ — OA and OB are both radii, so triangle OAB is isosceles and its base angles are equal.

3. 3

   $\angle OAT=90^{\circ }$ — The tangent is perpendicular to the radius at the point of contact.

4. 4

   $\angle TAB=90^{\circ }-24^{\circ }$ — TAB is the right angle OAT with the part OAB taken out of it.

5. 5

   $\angle TAB=66^{\circ }$ — Work out 90 − 24.

Worked example

1. 1

   $\angle ABC=56^{\circ }$ — Given. BA and BC are tangents from B, touching the circle at A and C.

2. 2

   $\angle OAB=\angle OCB=90^{\circ }$ — Each tangent is perpendicular to the radius at its point of contact.

3. 3

   $\angle AOC=360^{\circ }-90^{\circ }-90^{\circ }-56^{\circ }$ — OABC has four sides, so its angles add up to 360°. Take away the other three angles.

4. 4

   $\angle AOC=124^{\circ }$ — Work out 360 − 236.

5. 5

   $\angle ADC=124^{\circ }÷2=62^{\circ }$ — D is on the major arc, so the angle at the circumference is half the angle at the centre.

Worked example

1. 1

   $OP=13\text{ cm},OT=5\text{ cm}$ — OT is a radius. OP runs from the centre to the outside point P.

2. 2

   $\angle OTP=90^{\circ }$ — The tangent PT is perpendicular to the radius OT. So OTP is a right-angled triangle and OP, opposite the right angle, is the longest side.

3. 3

   $P{T}^2+5^2={13}^2$ — Pythagoras: the two shorter sides squared add up to the hypotenuse squared.

4. 4

   $P{T}^2=169-25=144$ — Take 5² away from both sides.

5. 5

   $PT=12\text{ cm}$ — Square root of 144.

Worked example

1. 1

   $PT=15\text{ cm},OT=8\text{ cm}$ — The tangent length and the radius. You want OP.

2. 2

   $O{P}^2=15^2+8^2$ — OP is the longest side, so add the squares of the two shorter sides.

3. 3

   $O{P}^2=225+64=289$ — Work out each square, then add.

4. 4

   $OP=17\text{ cm}$ — Square root of 289.

Worked example

1. 1

   $OP=6\text{ cm},\angle POQ=100^{\circ }$ — Tangents from R touch the circle at P and Q. You want the tangent length PR.

2. 2

   $\angle POR=100^{\circ }÷2=50^{\circ }$ — OR is the line of symmetry of the kite, so it cuts the angle at O in half.

3. 3

   $\angle OPR=90^{\circ }$ — The tangent is perpendicular to the radius, so triangle OPR has a right angle at P.

4. 4

   $\tan 50^{\circ }=\frac{PR}{6}$ — PR is opposite the 50° angle and OP = 6 is next to it. Opposite and next to, so use tan.

5. 5

   $PR=6\times \tan 50^{\circ }$ — Multiply both sides by 6.

6. 6

   $PR=7.15\text{ cm}$ — Use your calculator in degrees mode. Rounded to 3 significant figures.

Worked example

1. 1

   $\angle OPT=\angle OQT=90^{\circ }$ — Each tangent is perpendicular to the radius at its point of contact.

2. 2

   $OP=OQ$ — Both are radii of the same circle.

3. 3

   $OT\text{ is common}$ — OT is a side of both triangles, and it is the hypotenuse of each.

4. 4

   $△OPT\equiv △OQT$ — A right angle, the hypotenuse and one other side are equal in both. That makes the triangles congruent (RHS).

5. 5

   $TP=TQ$ — Matching sides of congruent triangles are equal.

Worked example

1. 1

   $\angle CBA=62^{\circ }$ — Given. AB is a diameter, DBE is the tangent at B, and C is on the circle on the same side as D.

2. 2

   $\angle ABD=90^{\circ }$ — The diameter passes through O, and the tangent is perpendicular to the radius.

3. 3

   $\angle DBC=90^{\circ }-62^{\circ }$ — DBC is the right angle ABD with the part CBA taken out.

4. 4

   $\angle DBC=28^{\circ }$ — Work out 90 − 62.

5. 5

   $\angle ACB=90^{\circ }$ — Angle in a semicircle, because AB is a diameter.

6. 6

   $\angle CAB=180^{\circ }-90^{\circ }-62^{\circ }=28^{\circ }$ — Angles in triangle ABC add up to 180°.

Worked example

1. 1

   $\angle AOC=360^{\circ }-90^{\circ }-90^{\circ }-70^{\circ }=110^{\circ }$ — Tangents from B with angle ABC = 70°. The kite OABC has two right angles, and its angles add up to 360°.

2. 2

   $\text{reflex }\angle AOC=360^{\circ }-110^{\circ }=250^{\circ }$ — The reflex angle goes the other way round O. Angles around a point add up to 360°.

3. 3

   $\angle ADC=110^{\circ }÷2=55^{\circ }$ — D is on the major arc, so the angle at the circumference is half the angle at the centre.

4. 4

   $\angle AEC=180^{\circ }-55^{\circ }=125^{\circ }$ — E is on the minor arc. A, D, C and E lie on the circle, so opposite angles of ADCE add up to 180°.