RONALD HUI TAK SUN SECONDARY SCHOOL HKDSE Mathematics RONALD HUI TAK SUN SECONDARY SCHOOL

Missing Homework SHW1-A1 (Re-do) SHW1-D1 SHW1-R1 SHW1-P1 12 9, 12 5J07, 9, 10, 12, 14, 19, 20, 23 SHW1-P1 23 Ronald HUI 2 October 2015

Missing Homework RE1 Summer Holiday Homework SQ1 5J07, 23, 24 Last Tuesday!!!! SQ1 Next Tuesday (6 Oct) Ronald HUI 2 October 2015

Tangents to a Circle and their Properties

If you draw a straight line, how many intersections do the line and the circle have? O

There are 3 possible cases. Q A B O x Q P T O x Q P O x 2 intersections 1 intersection no intersections PQ intersects the circle at two distinct points A and B. PQ intersects the circle at only one point T. PQ does not intersect the circle.

PQ touches the circle at T. In particular, in case 2, we say that O PQ touches the circle at T. P Q PQ is called the tangent to the circle at T. tangent T point of contact T is called the point of contact. For any point on the circle, we can draw one and only one tangent passing through it.

Radii and Tangents If PQ is the tangent to the circle at T, what is the relationship between the radius OT and the tangent PQ?

Consider a line PQ which cuts Radii and Tangents Consider a line PQ which cuts the circle at A and B. O P Q A B T

Radii and Tangents ∠OMB = 90 If PQ intersects OT at M such that AM = BM, then PQ⊥OT. ∠OMB = 90 O (line joining centre to mid-pt. of chord ⊥ chord) M P Q A B T

Radii and Tangents If we move PQ downwards such that the mid-point M always lie on OT, until A, M and B all coincide at T, O M P Q A B T

Radii and Tangents ∠OTQ = 90 If we move PQ downwards such that the mid-point M always lie on OT, until A, M and B all coincide at T, then PQ becomes the tangent to the circle at T. O ∠OTQ = 90 P Q T

Theorem 2.1 If PQ is the tangent to the circle at T, then PQ  OT. O P Abbreviation: tangent  radius O  P Q T

Let us use Theorem 2.1 to solve the following problem. In the figure, PQ is the tangent to the circle at T. Find x. Q P O T x 40 tangent  radius ∵ ÐOTQ = 90 ∴ x + 40 = 90 x = 50

In the figure, OT is a radius of the circle and PQ  OT. O  P Q T Is PQ the tangent to the circle at T? Yes, the converse of Theorem 2.1 is also true.

Theorem 2.2 (Converse of Theorem 2.1) Let OT be a radius of the circle and PTQ be a straight line. If PQ  OT, then PQ is the tangent to the circle at T. Abbreviation: converse of tangent  radius O  P Q T

Refer to the figure. Is AB the tangent to the circle at P? Example: Refer to the figure. Is AB the tangent to the circle at P? A B C D O P = 90 alt. s, AB // CD ∴ AB is the tangent to the circle at P. converse of tangent  radius

converse of tangent ⊥ radius Follow-up question In the figure, TP is the tangent to the circle at P. Prove that TQ is the tangent to the circle at Q. tangent  radius ÐOPT = 90 P In quadrilateral OQTP, ÐOQT + 50 + 90 + 130 = 360 O 130 ∠ sum of polygon ÐOQT = 90 50 ∴ TQ is the tangent to the circle at Q. T Q converse of tangent ⊥ radius

PQ is the tangent to the circle at T. PQ  OT In summary, we have: OT is a radius. Theorem 2.1 PQ is the tangent to the circle at T. PQ  OT Theorem 2.2 (converse of Theorem 2.1) O  P Q T

From Theorems 2.1 and 2.2, the theorem below follows directly. at its point of contact T The perpendicular to a tangent PQ passes through the centre O of the circle. P Q Abbreviation:  to tangent at its point of contact passes through centre O T

2 October 2015 Ronald HUI