Hello, and welcome to today's mathematics lesson. We are going to explore a beautiful chapter in geometry: Tangents and Intersecting Chords. By the end of this lesson, you will understand what makes a tangent special, how chords interact inside and outside circles, and some powerful theorems that connect angles, lengths, and circles in elegant ways.
Let us begin with something simple. Imagine a circle and a straight line drawn on the same plane. Exactly three things can happen between them.
First, the line might miss the circle entirely — they simply do not meet. Second, the line could cut through the circle, intersecting it at two distinct points. Such a line is called a secant of the circle.
Third, and most interestingly, the line might touch the circle at exactly one point. This line is called a tangent to the circle. That single point where they meet is known as the point of contact.
Now, here is our first fundamental theorem.
The tangent at any point of a circle and the radius through this point are perpendicular to each other.
Let me explain why this must be true. Consider a circle with centre O. Let AB be a tangent touching the circle at point P, and let OP be the radius to that point. We want to prove that OP is perpendicular to AB.
Take any other point Q on the tangent line, different from P, and join OQ. Since Q lies outside the circle, the distance OQ must be greater than the radius OP. Now, remember this key fact: of all line segments drawn from a given point to a given line, the perpendicular is the shortest. Since OP is the shortest distance from O to the line AB, it follows that OP must be perpendicular to AB.
This simple result has profound consequences. From it, we can deduce three important facts about tangents from external points.
First, no tangent can be drawn from a point inside the circle. Second, exactly one tangent can be drawn from a point on the circumference. Third, from any point outside the circle, exactly two tangents can be drawn.
And here is a beautiful corollary. If two tangents are drawn to a circle from an exterior point, then three remarkable things occur: the two tangents are equal in length; they subtend equal angles at the centre; and they are equally inclined to the line joining the external point to the centre.
The proof uses congruent triangles. Consider a circle with centre O, and let PA and PB be two tangents from external point P. In triangles AOP and BOP, we have OA = OB as radii, ∠OAP = ∠OBP = 90° by our tangent theorem, and OP is common. Therefore, the triangles are congruent by RHS. Hence PA = PB, ∠AOP = ∠BOP, and ∠APO = ∠BPO.
Now let us consider what happens when two circles touch each other.
If two circles touch each other, the point of contact lies on the straight line through the centres.
There are two cases: external touching and internal touching.
When two circles touch externally, draw the common tangent at the point of contact. Each radius to this point is perpendicular to the common tangent. Since both radii are perpendicular to the same line at the same point, they must lie on the same straight line. Therefore, the centres and the point of contact are collinear.
When two circles touch internally, the same logic applies. The radii to the common point of contact are both perpendicular to the common tangent, so they must coincide in direction, meaning the point of contact lies on the line joining the centres when extended.
This gives us useful formulas. If two circles with radii r₁ and r₂ touch externally, the distance between centres equals r₁ + r₂. If they touch internally, the distance equals the difference of the radii, r₁ − r₂ or r₂ − r₁, whichever is positive.
Let us work through a practical example.
Two circles with radii 25 centimetres and 9 centimetres touch each other externally. We want to find the length of their direct common tangent.
Imagine the two circles with centres A and B, touching at point O. Let MN be the direct common tangent, touching the larger circle at M and the smaller at N.
Draw BC perpendicular to AM. Then BCMN forms a rectangle, so BC = MN. Now AB = 25 + 9 = 34 cm, and AC = 25 − 9 = 16 cm, which is the difference of the radii.
Applying Pythagoras in triangle ABC: BC² = AB² − AC² = 34² − 16² = 1156 − 256 = 900. So BC = 30 cm, and therefore the direct common tangent has length 30 centimetres.
Now we come to one of the most elegant results in circle geometry: the alternate segment theorem.
The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
Imagine a circle with tangent PQ touching at point A, and a chord AB drawn through this point. The theorem states that the angle between tangent and chord, angle BAQ, equals the angle subtended by the chord in the alternate segment — that is, angle ACB where C is any point on the major arc.
To prove this, draw diameter AOR and join RB. Angle ABR is 90 degrees, being the angle in a semicircle. So angle ARB plus angle RAB equals 90 degrees. But angle RAB plus angle BAQ also equals 90 degrees, since the radius is perpendicular to the tangent. Therefore, angle ARB equals angle BAQ. Since angles ARB and ACB stand on the same arc, they are equal, proving that angle BAQ equals angle ACB.
The angle on the other side, angle BAP, similarly equals the angle in the opposite segment, angle ADB where D lies on the minor arc.
Now we turn to intersecting chords and the remarkable relationships they reveal.
If two chords of a circle intersect internally or externally, then the product of the lengths of their segments is equal.
Consider chords AB and CD intersecting at point P inside the circle. We prove that PA × PB = PC × PD.
Join AC and BD. In triangles APC and BPD, angle A equals angle D because they stand on the same arc BC. Similarly, angle C equals angle B as they stand on arc AD. Therefore, the triangles are similar by AA similarity. From the proportion of corresponding sides, PA/PD = PC/PB, which gives us PA × PB = PC × PD.
When chords intersect externally, the same result holds. Join AC and BD again. Now angle A equals angle PDB because the exterior angle of a cyclic quadrilateral equals the interior opposite angle. Similarly, angle C equals angle PBD. Triangles PAC and PDB are similar, leading to the same product equality.
Here is a special case of immense power.
If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord equals the square of the length of the tangent from the point of intersection to the point of contact.
Symbolically, if tangent PT and chord AB produced meet at P, then PA × PB = PT².
The proof uses the alternate segment theorem. Join TA and TB. In triangles PAT and PTB, angle PTB equals angle A by the alternate segment theorem, and angle P is common. Thus the triangles are similar, giving PA/PT = PT/PB, hence PT² = PA × PB.
Let us apply these theorems to solve a problem.
In triangle PQR, we have PQ = 24 cm, QR = 7 cm, and angle PQR equals 90 degrees. We need to find the radius of the inscribed circle.
First, by Pythagoras, PR equals √(24² + 7²), which equals √(576 + 49), which equals √625, giving 25 centimetres.
Let the inscribed circle touch PQ at A, QR at B, and PR at C. Since radii to points of contact are perpendicular to sides, and angle PQR is 90 degrees, the quadrilateral OAQB is a square with side equal to the radius x.
Therefore, AQ = BQ = x, so PA = 24 − x and RB = 7 − x. By the tangent property, PC = PA = 24 − x and RC = RB = 7 − x.
Since PR = PC + RC, we have 25 = (24 − x) + (7 − x), which simplifies to 25 = 31 − 2x. Solving, 2x = 6, so x = 3. The radius of the inscribed circle is 3 centimetres.
Let us try another application involving the alternate segment theorem.
Points A, B, and C lie on a circle. The tangent at C meets BA produced at T. Given that angle ATC equals 36 degrees and angle ACT equals 48 degrees, we want the angle subtended by AB at the centre.
First, in triangle ACT, the exterior angle CAB equals 48° + 36° = 84 degrees. Since the radius is perpendicular to the tangent, angle OCT equals 90 degrees, so angle OCA equals 90° − 48° = 42 degrees.
Since OA = OC, triangle OAC is isosceles, so angle OAC also equals 42 degrees. Therefore, angle OAB equals 84° − 42° = 42 degrees.
Similarly, OB = OA, so angle OBA equals 42 degrees. Finally, in triangle OAB, angle AOB equals 180° − 42° − 42° = 96 degrees. This is the angle subtended by chord AB at the centre.
Before we conclude, let us recall the key insights from this chapter.
First, a tangent is perpendicular to the radius at the point of contact — this is the foundation for all tangent properties.
Second, from any external point, two equal tangents can be drawn to a circle, and they make equal angles with the line to the centre.
Third, when two circles touch, their point of contact lies on the line through their centres, with distance between centres being the sum or difference of radii for external or internal touching.
Fourth, for intersecting chords, whether inside or outside the circle, the product of the segments of one chord equals the product of the segments of the other.
Fifth, when a tangent and a secant meet externally, the square of the tangent equals the product of the segments of the chord.
Sixth, the alternate segment theorem: the angle between a tangent and chord equals the angle in the alternate segment, a result of remarkable beauty and utility.
These theorems interconnect angles, lengths, and positions in ways that reveal the deep harmony of circle geometry. Practice visualising these configurations, and you will find that problems become puzzles waiting to be solved.
Thank you for your attention today. Keep exploring, keep questioning, and remember: every tangent touches a circle at just one point, but opens up infinite possibilities for understanding. Until next time, happy learning!