Tenth Grade Mathematics: Length of Tangents from an External Point Theorem
When an observer looks at a circular globe or planetary body from an external point, the lines of sight graze the curved perimeter at two symmetrical points of tangency. In Euclidean geometry, this visual symmetry is captured by a core theorem of tenth-grade geometry: the lengths of tangents drawn from an external point to a circle are strictly equal. This theorem, combined with its angular corollaries, provides the foundational mechanism for solving circumscribed polygons, calculating perimeter invariants, and analyzing cyclic symmetry.
Statement and Formal Proof of the Tangent Lengths Theorem
Theorem: The lengths of tangents drawn from an external point to a circle are equal.
Geometric Diagram for the Proof:
A (Point of contact)
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P + | r
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B (Point of contact)
Given: A circle with center O, an external point P, and two tangents PA and PB touching the circle at points A and B respectively. To Prove: PA = PB. Construction: Join the center O to the external point P (line segment OP), and join radii OA and OB.
Mathematical Proof
Consider the two triangles formed: triangle OAP and triangle OBP. By the Tangent-Radius Perpendicularity Theorem, the radius through the point of contact is perpendicular to the tangent:
Angle OAP = 90 degrees and Angle OBP = 90 degrees
Thus, both triangle OAP and triangle OBP are right-angled triangles.
Now compare their corresponding sides:
1. Hypotenuse OP is common to both triangles (OP = OP).
2. Leg OA is equal to leg OB because both are radii of the exact same circle (OA = OB = r).
Therefore, by the RHS (Right angle - Hypotenuse - Side) Congruence Criterion:
triangle OAP ~= triangle OBP
Because corresponding parts of congruent triangles (CPCT) are strictly equal:
PA = PB
This completes the formal proof that the lengths of tangents from an external point are equal.
Angular Corollaries of the Congruence
From the identical triangle congruence (triangle OAP ~= triangle OBP), two vital angular properties follow immediately:
1. Tangents Subtend Equal Angles at the Center:
Angle AOP = Angle BOP
The line joining the external point to the center bisects the angle subtended
by the points of contact at the center.
2. Line from External Point Bisects the Tangent Angle:
Angle APO = Angle BPO
The line joining the external point to the center bisects the angle formed
between the two tangents.
3. Supplementary Angle Property:
In quadrilateral OAPB:
Angle APB + Angle AOB = 180 degrees
The angle between the two tangents and the angle subtended by the line segment
joining the points of contact at the center are supplementary.
Circumscribed Polygons and Perimeter Invariants
When a polygon circumscribes a circle (meaning every side of the polygon is tangent to the circle), every vertex acts as an external point from which two equal tangent segments emanate.
Master Circumscribed Quadrilateral Theorem:
If a quadrilateral ABCD circumscribes a circle, then the sum of opposite sides
is equal:
AB + CD = AD + BC
Proof:
Let the circle touch sides AB, BC, CD, DA at points P, Q, R, S respectively.
By the equal tangent lengths theorem:
AP = AS --- (1) (Tangents from vertex A)
BP = BQ --- (2) (Tangents from vertex B)
CR = CQ --- (3) (Tangents from vertex C)
DR = DS --- (4) (Tangents from vertex D)
Add equations (1), (2), (3), (4):
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
Notice that AP + BP = AB, CR + DR = CD, AS + DS = AD, and BQ + CQ = BC.
Therefore:
AB + CD = AD + BC
Corollary for Parallelograms:
If a parallelogram circumscribes a circle, it must be a RHOMBUS!
Proof: In a parallelogram, opposite sides are equal: AB = CD and AD = BC.
Substitute into AB + CD = AD + BC:
AB + AB = AD + AD ===> 2*AB = 2*AD ===> AB = AD.
Since adjacent sides are equal, all four sides are equal, proving it is a rhombus.
Chapter Exercises
Exercise 1: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. Find the radius of the circle.
Exercise 2: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.
Exercise 3: A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC.
Solutions and Explanations
Solution 1: Let O be the center of the circle, and let P be the point of contact of the tangent from Q. The tangent length is QP = 24 cm, and the distance from the center is OQ = 25 cm. By the tangent-radius theorem, OP is perpendicular to QP: Angle OPQ = 90 degrees. In right-angled triangle OPQ, by the Pythagorean theorem: OQ^2 = OP^2 + QP^2 ==> 25^2 = OP^2 + 24^2 ==> 625 = OP^2 + 576 ==> OP^2 = 625 - 576 = 49 ==> OP = sqrt(49) = 7 cm. Thus, the radius of the circle is 7 cm.
Solution 2: Let P be an external point and let PA and PB be tangents drawn to the circle with center O, touching at A and B. Radii OA and OB meet the tangents at right angles: Angle OAP = 90 degrees and Angle OBP = 90 degrees. Consider the quadrilateral OAPB. The sum of the interior angles of any quadrilateral is 360 degrees: Angle APB + Angle OAP + Angle AOB + Angle OBP = 360 ==> Angle APB + 90 + Angle AOB + 90 = 360 ==> Angle APB + Angle AOB + 180 = 360 ==> Angle APB + Angle AOB = 360 - 180 = 180 degrees. Since their sum is 180 degrees, the angle between the two tangents (Angle APB) and the angle subtended by the points of contact at the center (Angle AOB) are supplementary.
Solution 3: Let the circle touch sides BC, CA, and AB at points D, E, and F respectively. The radius is r = 4 cm. Given BD = 8 cm and DC = 6 cm. By the equal tangents theorem: CF = CD = 6 cm, and BE = BD = 8 cm (tangents from C and B). For vertex A, let the equal tangent lengths be AE = AF = x. The sides of triangle ABC are: a = BC = 8 + 6 = 14 cm; b = AC = x + 6 cm; and c = AB = x + 8 cm. The semi-perimeter s is: s = (a + b + c)/2 = (14 + x + 6 + x + 8)/2 = (2x + 28)/2 = x + 14. We can calculate the area of triangle ABC in two ways. First, using Heron's formula: Area = sqrt(s * (s - a) * (s - b) * (s - c)) = sqrt((x + 14) * (x + 14 - 14) * (x + 14 - (x + 6)) * (x + 14 - (x + 8))) = sqrt((x + 14) * x * 8 * 6) = sqrt(48 * x * (x + 14)). Second, decomposing triangle ABC into three triangles OBC, OCA, and OAB using the inradius r = 4 cm: Area = Area(OBC) + Area(OCA) + Area(OAB) = (1/2)ar + (1/2)br + (1/2)cr = (1/2)r(a + b + c) = r * s = 4 * (x + 14). Equating both area expressions: sqrt(48 * x * (x + 14)) = 4 * (x + 14). Square both sides: 48 * x * (x + 14) = 16 * (x + 14)^2. Divide both sides by 16(x + 14) (valid since x + 14 != 0): 3x = x + 14 ==> 2x = 14 ==> x = 7 cm. Now find the side lengths: AB = c = x + 8 = 7 + 8 = 15 cm. AC = b = x + 6 = 7 + 6 = 13 cm. Thus, side AB is 15 cm and side AC is 13 cm.