Three guided lessons with explanations, visual steps, practice and a final check.
Original learning companion, not full textbook exercise solutions. Covers the radius–tangent relation, equal tangent lengths with RHS reasoning, and a tangent-angle consequence. It provides a concise proof explanation, not every textbook proof or exercise configuration.
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LESSON 1
Locate the radius perpendicular to a tangent
Identify a tangent and apply the right angle at its contact point.
A tangent touches a circle at one point; a secant intersects it at two points.
Identify the contact
The line touches the circle at T. Join the centre O to T to draw a radius.
Use the theorem
The radius OT is perpendicular to the tangent at T, so the contact angle is 90°.
Understand why the foot matters
The perpendicular from O to a line is its shortest distance. If a different point on a tangent were closer than T, it would lie inside the circle and the line would cut the circle.
Use the right angle in problems
When an external point P lies on the tangent, △OTP is right-angled at T. Pythagoras and triangle-angle rules apply.
Worked example: A circle has radius 5 cm and OP = 13 cm for an external point P with tangent PT. Since OT ⟂ PT, PT = √(13² − 5²) = √144 = 12 cm.
Watch for this: A line perpendicular to a radius is the tangent only when it passes through that radius endpoint on the circle.
Discuss: Draw a tangent at the top of a circle. Show which radius creates the right angle and distinguish a parallel line that misses the circle.
LESSON 2
Compare tangents from one external point
Use equal tangent lengths and the congruent-triangle reason.
Radii of the same circle have equal lengths.
Name both contacts
PA and PB are tangents from the same outside point P. Join O to A, O to B and O to P.
Find equal parts
OA = OB because they are radii. OP is common, and both contact angles are 90°.
Apply RHS congruence
Right triangles OAP and OBP are congruent by hypotenuse OP and equal radius sides. Therefore PA = PB.
Use the equality
If PA = 2x + 1 and PB = x + 6, then 2x + 1 = x + 6, so x = 5 and both tangent lengths are 11.
Worked example: If PA = 3x − 2 and PB = x + 8, equality gives 2x = 10 and x = 5. Both lengths equal 13 units, which are positive as required.
Watch for this: Tangent lengths are equal from the same external point. Tangents from different external points need not have equal lengths.
Discuss: Sketch two different outside points. Explain why the theorem pairs two tangent lengths at each point, not all four lengths together.
LESSON 3
Connect central and tangent angles
Combine two tangent right angles with an angle sum.
The interior angles of a quadrilateral total 360°.
Add both radii
For tangents PA and PB, join OA and OB. This creates quadrilateral OAPB.
Mark the contact angles
∠OAP = 90° and ∠OBP = 90°. Together they contribute 180°.
Use the quadrilateral sum
Thus ∠AOB + ∠APB = 180° when AOB is the non-reflex angle inside this quadrilateral.
Calculate a concrete angle
If ∠AOB = 120°, then ∠APB = 180° − 120° = 60°. Check all four angles add to 360°.
Worked example: If the angle between two tangents is 50°, then the relevant central angle is 180° − 50° = 130°. The four interior angles are 90°, 90°, 50° and 130°.
Watch for this: Do not subtract a reflex central angle from 180°. Identify the angle inside the quadrilateral.
Discuss: Choose three possible central angles between 0° and 180°. Calculate the tangent angle and describe how the two change together.
Original learning companion, not full textbook exercise solutions. Covers the radius–tangent relation, equal tangent lengths with RHS reasoning, and a tangent-angle consequence. It provides a concise proof explanation, not every textbook proof or exercise configuration.
Original HelioLearn teaching and questions. Source-aligned companion; not an official NCERT/CBSE publication or full textbook solution bank. Independent teacher review and narrated video are pending.
