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SPI Mathematics · Circles

Tangents from an External Point

The two tangents from a point outside a circle are equal, and the angle between them and the angle between the radii add to 180 degrees.

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Why this matters

Twelve questions in the bank, five EASY and seven MODERATE: a third tangent cutting a pair of tangents, the angle between two tangents, a square of tangents and radii, and the circle inside a right triangle. One idea settles all of them: from any outside point, the two tangent pieces are equal.

Concept 1 of 3: Two tangents from one point are equal

Draw tangents PA and PB from an outside point P to a circle with centre O. Triangles OAP and OBP are both right-angled, share the side OP and have equal radii OA and OB, so they are congruent. That makes PA equal to PB.

Definition

  • PA=PBPA = PB: the two tangents from an outside point are equal.
  • OP bisects ∠APB\angle APB, so ∠APO=∠BPO\angle APO = \angle BPO.
  • Every outside point gives its own equal pair. When a third tangent cuts the first two, split each long tangent into pieces and match the equal pairs.
  • If PA=PBPA = PB and ∠APB=60∘\angle APB = 60^\circ, triangle PAB is equilateral, so AB = PA.
  • A circle inside a quadrilateral ABCD that touches all four sides gives AB+CD=AD+BCAB + CD = AD + BC.

Equal tangents

PA=PBAB+CD=AD+BC (circle touching all four sides)PA = PB \qquad AB + CD = AD + BC \text{ (circle touching all four sides)}

Worked example

From a point P, tangents PA and PB are drawn to a circle. A third tangent touches the circle at R and meets PA at C and PB at D. If PA = 10 cm, find the perimeter of triangle PCD.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Circles · Tangents from an External Point

CP and CQ are tangents from the point C to the circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm. and BR = 4 cm. Then length of BC =

Matching the wrong pair of pieces

When B lies on the tangent CQ and BR is a tangent from B, the equal pair is BR and BQ. Find BQ as CQ minus CB first; BR is not equal to CB.

Thinking only the first point gives equal tangents

Every outside point has two equal tangents, including the points where a third tangent cuts the first two. Use each of them.
Drill 6 more on two tangents from one point are equal

Concept 2 of 3: The angle between two tangents and the angle at the centre

The two tangents and the two radii make a four-sided figure. Its angles add to 360 degrees and two of them are right angles, so the angle between the tangents and the angle between the radii share the other 180 degrees.

Definition

  • ∠OAP=∠OBP=90∘\angle OAP = \angle OBP = 90^\circ, so in quadrilateral OAPB, ∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^\circ.
  • To get tangents at a chosen angle, draw radii at 180∘180^\circ minus that angle.
  • OA = OB, so triangle OAB is isosceles and ∠OAB=∠OBA=12∠APB\angle OAB = \angle OBA = \frac{1}{2}\angle APB.
  • If ∠APB=90∘\angle APB = 90^\circ, OAPB is a square of side rr, and each diagonal (AB and OP) is r2r\sqrt{2}.

Tangents and radii

∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^\circ

Worked example

Two tangents to a circle meet at P at an angle of 70∘70^\circ. Find the angle between the radii drawn to the two points of contact.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Circles · Tangents from an External Point

To draw a pair of tangents to a circle which are inclined to each other at an angle of 35∘35^\circ, it is required to draw tangents at the end-points of those two radii of the circle, the angle between which is

Taking 90 minus the angle instead of 180 minus it

Tangents at 40∘40^\circ need radii at 140∘140^\circ, not 50∘50^\circ. The two angles add to 180∘180^\circ because the other two corners are right angles.

Forgetting the two right angles

The angles of OAPB add to 360∘360^\circ, and the corners at the points of contact are 90∘90^\circ each. Subtract both before finding the last angle.
Drill 3 more on the angle between two tangents and the angle at the centre

Concept 3 of 3: The circle inside a triangle

The circle inside a triangle touches all three sides, so each corner is an outside point with two equal tangents. In a right triangle the corner at the right angle makes a small square with the centre, which gives the radius at once.

Definition

  • The incircle touches all three sides. From each corner the two tangent pieces are equal.
  • In a right triangle with legs aa, bb and hypotenuse cc, the tangents from the right-angle corner are rr each. The other pieces are a−ra - r and b−rb - r, and they make up the hypotenuse: c=(a−r)+(b−r)c = (a - r) + (b - r).
  • For any triangle: r=Areasr = \dfrac{\text{Area}}{s}, where ss is half the perimeter.
  • The perimeter is twice the sum of the three different tangent lengths.

Radius of the incircle

r=a+b−c2 (right triangle)r=Areas,s=perimeter2r = \frac{a + b - c}{2} \text{ (right triangle)} \qquad r = \frac{\text{Area}}{s}, \quad s = \frac{\text{perimeter}}{2}

Worked example

A right triangle has legs 8 cm and 15 cm. Find the radius of its incircle.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Circles · Tangents from an External Point

In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is

Dividing the area by the whole perimeter

In r=Areasr = \frac{\text{Area}}{s}, ss is half the perimeter. Dividing by the whole perimeter gives half the true radius.

Halving the hypotenuse

Half the hypotenuse of a right triangle is the radius of the circle through its three corners, a different circle. The circle inside has radius a+b−c2\frac{a + b - c}{2}.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Two tangents from one point are equal

    Equal tangents

    PA=PBAB+CD=AD+BC (circle touching all four sides)PA = PB \qquad AB + CD = AD + BC \text{ (circle touching all four sides)}
  • The angle between two tangents and the angle at the centre

    Tangents and radii

    ∠APB+∠AOB=180∘\angle APB + \angle AOB = 180^\circ
  • The circle inside a triangle

    Radius of the incircle

    r=a+b−c2 (right triangle)r=Areas,s=perimeter2r = \frac{a + b - c}{2} \text{ (right triangle)} \qquad r = \frac{\text{Area}}{s}, \quad s = \frac{\text{perimeter}}{2}

Watch out for (6)

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