MathsCBSE Class 10 · Circles

Lengths of Tangents from an External Point: Proof and Problems

Updated 14 October 2026 · Fact-checked

The two tangents drawn from a point outside a circle to that circle are always equal in length. This is Theorem 10.2 of NCERT Class 10 Circles, proved using RHS congruence. The board asks you to prove it, often in Section C or D, and to use it to find unknown lengths, angles and perimeters.

Radii OA and OB meet the tangents at 90°, and OP is shared, so triangles OAP and OBP are congruent by RHS and PA = PB.

Illustration drawn for QuestionBuddy.

Understand Lengths of Tangents from an External Point

Take a circle with centre $O$ and a point $P$ outside it. From $P$ you can draw exactly two tangents, touching the circle at $A$ and $B$ . The length of a tangent means the length of the segment from the external point to the point of contact, so here it is $P A$ or $P B$ . A tangent line is infinite; its length only makes sense as this segment.

Theorem 10.2 says $P A = P B$ . Why? Join $O A$ , $O B$ and $O P$ . By Theorem 10.1 the radius is perpendicular to the tangent, so $∠ O A P = ∠ O B P = 9 0^{\circ}$ . Also $O A = O B$ because both are radii, and $O P$ is common. Two right triangles with equal hypotenuses and one equal side are congruent by RHS. So $△ O A P ≅ △ O B P$ , and $P A = P B$ by CPCT.

The same congruence gives extra results for free. $∠ A P O = ∠ B P O$ , so $O P$ bisects the angle between the tangents. $∠ A O P = ∠ B O P$ , so $O P$ also bisects the angle $∠ A O B$ at the centre. Since $P$ and $O$ are each equidistant from $A$ and $B$ , $O P$ is the perpendicular bisector of the chord $A B$ . And because $P A = P B$ , $△ P A B$ is isosceles.

You can also see it with Pythagoras: $P A^{2} = O P^{2} - O A^{2}$ and $P B^{2} = O P^{2} - O B^{2}$ . The right-hand sides are equal, so $P A = P B$ . This is a good check, but in the board proof RHS congruence is the expected route.

The result holds only when the point is outside the circle. A point on the circle has one tangent, and a point inside has none. The theorem is the engine of the whole chapter: every triangle or quadrilateral drawn around a circle is solved by marking equal tangents from each vertex, and every perimeter question with a third tangent uses it repeatedly.

Lengths of Tangents from an External Point: formulas to remember

Theorem 10.2: equal tangents: $P A = P B$; PA and PB are tangents from the same external point P. Proof: RHS congruence of triangles OAP and OBP.
Tangent length by Pythagoras: $P A = P B = O P^{2} - r^{2}$; OP is the hypotenuse. Give the reason (radius perpendicular to tangent) before using it.
OP bisects both angles: $∠ A P O = ∠ B P O = \frac{1}{2} ∠ A P B, ∠ A O P = ∠ B O P$; From the same congruence, by CPCT.
OP is perpendicular bisector of AB: $O P ⊥ A B, A M = M B$; M is the point where OP meets AB.
Base angles of triangle PAB: $∠ P A B = ∠ P B A = 9 0^{\circ} - \frac{1}{2} ∠ A P B$; Triangle PAB is isosceles because PA = PB.
Third tangent perimeter: $P C + C D + D P = 2 P A$; CD touches the circle between the two tangents, with C on PA and D on PB.
Angle at P and angle at O: $∠ A P B + ∠ A O B = 18 0^{\circ}$; In quadrilateral OAPB the angles at A and B are each $9 0^{\circ}$ , so the other two add to $18 0^{\circ}$ .
Quadrilateral circumscribing a circle: $A B + C D = A D + B C$; Mark the four pairs of equal tangents from A, B, C, D and add; the opposite sides have equal sums.

How to solve Lengths of Tangents from an External Point questions

Use this routine on any question about tangents from an external point, whether it asks for a length, an angle or a proof.

1Describe or sketch the figure: centre, external point, both points of contact.
2Join the centre to each point of contact and to the external point, and mark both right angles (Theorem 10.1).
3Mark equal tangents from every external point with the same letter or expression.
4If the tangents are given as algebraic expressions, equate them and solve for the unknown first.
5For lengths, use the right triangle with $O P$ as hypotenuse; for angles, use the isosceles triangle $P A B$ or the fact that $O P$ bisects the angle.
6For perimeters, split each tangent at its point of contact and replace every piece by its equal tangent.
7Write a reason in brackets on every line and check the answer by substitution.

Lengths of Tangents from an External Point in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Theorem 10.2: tangents from an external point are equal

Imagine a kabaddi coach ties a rope from a pole P to a circular water tank of radius 9 m. Two straight ropes leave P and just touch the tank, at A and B. The pole is 41 m from the centre of the tank.

Find the length of each rope, the total rope used if both ropes run from P to their points of contact, and the straight-line distance AB between the two points of contact.

Try it first, then show the solutionSolution

Radius $O A ⊥ P A$ (tangent is perpendicular to radius), so $△ O A P$ is right-angled at $A$ .
$P A^{2} = O P^{2} - O A^{2} = 4 1^{2} - 9^{2} = 1681 - 81 = 1600$ , so $P A = 40$ m.
$P B = P A = 40$ m (tangents from an external point are equal), so the total rope is 80 m.
$O P$ is the perpendicular bisector of $A B$ ; let it meet $A B$ at $M$ . Area of $△ O A P$ : $\frac{1}{2} \times 9 \times 40 = \frac{1}{2} \times 41 \times A M$ , so $A M = \frac{360}{41}$ m.
$A B = 2 A M = \frac{720}{41} \approx 17.56$ m.

Answer: Each rope is 40 m; total 80 m; the points of contact are about 17.56 m apart ( $\frac{720}{41}$ m).

Memory hook: Two ropes from one pole to a round tank are always twins.

Third tangent perimeter

Suppose a school sports day track has a circular stage. From the gate P, two straight fences run and just touch the stage at A and B, each 15 m long. A banner rod CD touches the stage between them, with C on fence PA and D on fence PB.

The banner rod CD is 10 m long, and C is 3 m nearer to the gate P than D is. Find PC, PD, CA and DB, and the perimeter of triangle PCD.

Try it first, then show the solutionSolution

Let the rod touch the stage at E. Then $C E = C A$ and $D E = D B$ (equal tangents), so the perimeter of $△ P C D = P C + C A + D B + P D = P A + P B = 30$ m.
So $P C + P D = 30 - C D = 20$ m.
With $P D = P C + 3$ : $2 P C + 3 = 20$ , so $P C = 8.5$ m and $P D = 11.5$ m.
$C A = 15 - 8.5 = 6.5$ m and $D B = 15 - 11.5 = 3.5$ m.
Check: $C D = C E + E D = C A + D B = 6.5 + 3.5 = 10$ m.

Answer: PC = 8.5 m, PD = 11.5 m, CA = 6.5 m, DB = 3.5 m; perimeter of triangle PCD = 30 m.

Memory hook: Cut the rod at its touch point and each half slides back onto a fence.

Tricks to remember Lengths of Tangents from an External Point

Short memory aids. Say each one aloud once, then use it on a question the same day.

Proof in three letters: R, H, S
Right angle (tangent ⊥ radius), Hypotenuse common ( $O P$ ), Side equal ( $O A = O B$ , radii). Then CPCT.
Use it when: Write the three pairs in this order in every proof of Theorem 10.2.
One point, one length
Every external point owns one tangent length. Give it one letter and use it on both sides.
Use it when: Triangles or quadrilaterals around a circle, and third-tangent perimeters.
OP is the mirror
The figure is symmetric about $O P$ : it halves the angle at $P$ , halves the angle at $O$ and cuts $A B$ at right angles in the middle.
Use it when: Angle questions and finding the chord of contact.

Lengths of Tangents from an External Point: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Equal tangents with algebra · 2 marks · Very short answerBoard level

$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ . $P A = (3 x + 2)$ cm and $P B = (5 x - 6)$ cm. If the radius is $5$ cm, find $x$ , $P A$ and $O P$ .

Try it first, then show the solutionSolution

$P A = P B$ (tangents from an external point are equal), so $3 x + 2 = 5 x - 6$ , giving $x = 4$ . (½)
$P A = 3 (4) + 2 = 14$ cm. (½)
$∠ O A P = 9 0^{\circ}$ (radius ⊥ tangent), so $O P^{2} = 1 4^{2} + 5^{2} = 196 + 25 = 221$ . (½)
$O P = 221$ cm. (½)

Answer: $x = 4$ , $P A = 14$ cm, $O P = 221$ cm.

Example 2: Prove Theorem 10.2 and find the chord · 5 marks · Long answerHarder board level

(i) Prove that the lengths of tangents drawn from an external point to a circle are equal.
(ii) Tangents $P A$ and $P B$ are drawn from $P$ to a circle with centre $O$ and radius $6$ cm, with $O P = 10$ cm. Find $P A$ and the length of chord $A B$ .

Try it first, then show the solutionSolution

Given: circle with centre $O$ , tangents $P A$ , $P B$ from external point $P$ . To prove: $P A = P B$ . Construction: join $O A$ , $O B$ , $O P$ . (½)
$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent ⊥ radius at point of contact). (½)
$O A = O B$ (radii) and $O P = O P$ (common). (½)
$△ O A P ≅ △ O B P$ (RHS), so $P A = P B$ (CPCT). (½)
(ii) $P A^{2} = O P^{2} - O A^{2} = 100 - 36 = 64$ , so $P A = 8$ cm. (1)
$O P ⊥ A B$ and bisects it, because $O A = O B$ and $P A = P B$ make $O P$ the perpendicular bisector of $A B$ ; let it meet $A B$ at $M$ . Area of $△ O A P$ : $\frac{1}{2} (O A) (P A) = \frac{1}{2} (O P) (A M)$ . (½)
$6 \times 8 = 10 \times A M$ , so $A M = 4.8$ cm. (½)
$A B = 2 A M = 9.6$ cm. (1)

Answer: $P A = P B$ proved; $P A = 8$ cm, $A B = 9.6$ cm.

Example 3: Angle between tangents and the chord · 3 marks · Short answerBoard level

$P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ . If $∠ A P B = 7 0^{\circ}$ , find $∠ P A B$ , $∠ A O B$ and $∠ O A B$ , giving reasons.

Try it first, then show the solutionSolution

$P A = P B$ (equal tangents), so $∠ P A B = ∠ P B A$ . (½)
$∠ P A B = \frac{18 0 ^{\circ} - 7 0 ^{\circ}}{2} = 5 5^{\circ}$ (angle sum of triangle). (½)
In quadrilateral $O A P B$ , $∠ O A P = ∠ O B P = 9 0^{\circ}$ (radius ⊥ tangent). (½)
$∠ A O B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ . (½)
$∠ O A B = ∠ O A P - ∠ P A B = 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ . (½)
Check: $O A = O B$ , so $∠ O A B = \frac{18 0 ^{\circ} - 11 0 ^{\circ}}{2} = 3 5^{\circ}$ . (½)

Answer: $∠ P A B = 5 5^{\circ}$ , $∠ A O B = 11 0^{\circ}$ , $∠ O A B = 3 5^{\circ}$ .

Example 4: Two circles and a common external point · 3 marks · Short answerHarder board level

Two circles touch each other externally at $T$ . A line through $T$ is the common tangent at $T$ . $P$ is a point on this line other than $T$ . From $P$ , tangent $P A$ is drawn to the first circle (touching at $A$ ) and tangent $P B$ to the second circle (touching at $B$ ). Prove that $P A = P B$ . If $P T = 7$ cm, what are $P A$ and $P B$ ?

Try it first, then show the solutionSolution

$P$ is external to the first circle, and $P A$ and $P T$ are both tangents to it from $P$ . (½)
So $P A = P T$ (equal tangents from an external point). (1)
Similarly $P B$ and $P T$ are tangents from $P$ to the second circle, so $P B = P T$ . (1)
Hence $P A = P B$ , and $P A = P B = 7$ cm. (½)

Answer: $P A = P B = P T = 7$ cm.

Example 5: Case study: the bangle on a table · 4 marks · Case-basedHOTS / competency

Suppose a bangle of radius $3$ cm lies flat on a table with centre $O$ . Two straight sticks start at a point $P$ and just touch the bangle at $A$ and $B$ . A ruler is laid to touch the bangle at $E$ on the arc between them, meeting $P A$ at $C$ and $P B$ at $D$ . $O P = 5$ cm.
(i) Find $P A$ . (1)
(ii) Find the length of the chord $A B$ joining the two points of contact. (1)
(iii) Find the perimeter of $△ P C D$ . If $P C = 2.5$ cm, find $C E$ . (2)

Try it first, then show the solutionSolution

(i) $P A^{2} = O P^{2} - O A^{2} = 25 - 9 = 16$ (radius ⊥ tangent), so $P A = 4$ cm. (1)
(ii) $O P$ is the perpendicular bisector of $A B$ ; let it meet $A B$ at $M$ . Area of $△ O A P$ : $\frac{1}{2} \times O A \times P A = \frac{1}{2} \times O P \times A M$ , so $3 \times 4 = 5 \times A M$ , $A M = 2.4$ cm and $A B = 4.8$ cm. (1)
(iii) $C E = C A$ and $D E = D B$ (equal tangents), so perimeter $= P C + C A + D B + P D = P A + P B = 8$ cm. (1)
$C E = C A = P A - P C = 4 - 2.5 = 1.5$ cm. (1)

Answer: (i) $4$ cm (ii) $A B = 4.8$ cm (iii) perimeter $8$ cm, $C E = 1.5$ cm.

Example 6: Opposite sides of a circumscribed quadrilateral · 3 marks · Short answerHarder board level

A quadrilateral ABCD is drawn so that each of its sides touches a circle: AB at P, BC at Q, CD at R and DA at S. Prove that $A B + C D = A D + B C$ . If $A B = 6$ cm, $B C = 7$ cm and $C D = 4$ cm, find $A D$ .

Try it first, then show the solutionSolution

Tangents from an external point are equal: $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$ . (1 mark)
Adding: $A P + B P + C R + D R = A S + B Q + C Q + D S$ , that is $(A P + P B) + (C R + R D) = (A S + S D) + (B Q + QC)$ . (½ mark)
So $A B + C D = A D + B C$ . (½ mark)
$6 + 4 = A D + 7$ , so $A D = 3$ cm. (1 mark)

Answer: $A B + C D = A D + B C$ proved; $A D = 3$ cm.

Where students lose marks in Lengths of Tangents from an External Point

The slips examiners see most often, and what to write instead.

Mistake: Missing the reason for the right angle in the proof
RHS needs a right angle, and the examiner wants to see where it comes from.
Wrong working
$∠ O A P = ∠ O B P = 9 0^{\circ}$ .
Right working
$∠ O A P = ∠ O B P = 9 0^{\circ}$ (tangent at a point is perpendicular to the radius through the point of contact).
Do this: Quote Theorem 10.1 in brackets every time.
Usually costs: ½ to 1 mark
Mistake: Using SAS or ASA instead of RHS
You only know a right angle, the hypotenuse and one side; the included angle is not known.
Wrong working
$△ O A P ≅ △ O B P$ (SAS).
Right working
$△ O A P ≅ △ O B P$ (RHS: right angle, hypotenuse $O P$ common, side $O A = O B$ ).
Do this: Match the three pairs to R, H, S explicitly.
Usually costs: 1 mark
Mistake: Treating $O P$ as a leg
The right angle is at the point of contact, so $O P$ is opposite it.
Wrong working
$P A^{2} = O P^{2} + r^{2}$ .
Right working
$P A^{2} = O P^{2} - r^{2}$ .
Do this: Mark the right angle at $A$ first; the side opposite it is the hypotenuse.
Usually costs: Full marks for the part
Mistake: Counting the third tangent as a whole in the perimeter
Students add $C D$ but forget to split it at the contact point.
Wrong working
Perimeter $= P A + P B + C D$ .
Right working
$C D = C A + D B$ , so perimeter $= P C + C A + D B + P D = P A + P B$ .
Do this: Always split a tangent at its point of contact.
Usually costs: 1 to 2 marks

Exam tips

Write Given, To prove, Construction and Proof for Theorem 10.2; a described figure counts.
Before any calculation, list every pair of equal tangents the figure gives you.
In angle MCQs, check your answer two ways: through triangle $P A B$ and through quadrilateral $O A P B$ .
In assertion-reason items, check that the reason (for example, tangents from an external point are equal) really explains the assertion, and remember it applies only to a point outside the circle.
State units in every final length.

Practice questions on Lengths of Tangents from an External Point

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ . If $∠ A P B = 8 0^{\circ}$ , then $∠ P O A$ is
(a) $4 0^{\circ}$
(b) $5 0^{\circ}$
(c) $8 0^{\circ}$
(d) $10 0^{\circ}$
Hint
$O P$ bisects $∠ A P B$ ; use the right angle at $A$ .
Answer
(b) $5 0^{\circ}$ . $∠ A P O = 4 0^{\circ}$ , so $∠ P O A = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ .
Q2 · 3 marks · Short answerHarder board level
Prove that the line joining the external point $P$ to the centre $O$ is the perpendicular bisector of the chord $A B$ joining the points of contact of tangents $P A$ and $P B$ .
Hint
First get $∠ A P O = ∠ B P O$ , then compare triangles $P A M$ and $P B M$ where $M$ is on $A B$ .
Answer
$△ O A P ≅ △ O B P$ (RHS) gives $∠ A P M = ∠ B P M$ . In $△ P A M$ and $△ P B M$ : $P A = P B$ , $∠ A P M = ∠ B P M$ , $P M$ common, so congruent by SAS. Hence $A M = B M$ and $∠ A M P = ∠ B M P$ ; these are a linear pair, so each is $9 0^{\circ}$ . So $O P$ is the perpendicular bisector of $A B$ .
Q3 · 3 marks · Short answerHarder board level
From a point $P$ , $26$ cm from the centre of a circle of radius $10$ cm, tangents $P A$ and $P B$ are drawn. A third tangent touching the circle at a point on the minor arc $A B$ meets $P A$ at $C$ and $P B$ at $D$ . Find the perimeter of $△ P C D$ .
Hint
Find $P A$ by Pythagoras, then replace $C E$ and $E D$ by equal tangents.
Answer
$P A = 676 - 100 = 24$ cm. Perimeter $= P C + C E + E D + D P = P C + C A + D B + D P = P A + P B = 48$ cm.
Q4 · 3 marks · Short answerBoard level
Two tangents $P A$ and $P B$ from $P$ to a circle with centre $O$ are perpendicular to each other. If the radius is $7$ cm, find $P A$ and $O P$ , and name the shape $O A P B$ with reasons.
Hint
Three angles of $O A P B$ are right angles; check adjacent sides.
Answer
$∠ O A P = ∠ O B P = 9 0^{\circ}$ (radius ⟂ tangent) and $∠ A P B = 9 0^{\circ}$ (given), so $∠ A O B = 36 0^{\circ} - 27 0^{\circ} = 9 0^{\circ}$ and OAPB is a rectangle. Its adjacent sides OA and OB are equal radii (7 cm), so it is a square. Hence $P A = P B = 7$ cm, and OP, its diagonal, is $72$ cm.
Q5 · 2 marks · Very short answerBoard level
Imagine a cycle chain modelled as two straight pieces $P A$ and $P B$ leaving a point $P$ and touching a circular gear. A student says: " $P A = 12$ cm and $O P = 13$ cm, so the radius is $1 2^{2} + 1 3^{2}$ cm." Find the error and the correct radius, and find $P B$ .
Hint
Which side is the hypotenuse in $△ O A P$ ?
Answer
$O P$ is the hypotenuse (right angle at $A$ ), so $r^{2} = 1 3^{2} - 1 2^{2} = 25$ , $r = 5$ cm. The student added instead of subtracting. $P B = P A = 12$ cm.
Q6 · 3 marks · Short answerHarder board level
A circle is inscribed in $△ A B C$ , touching $A B$ at $F$ , $B C$ at $D$ and $C A$ at $E$ . If $A B = 12$ cm, $B C = 8$ cm and $C A = 10$ cm, find $A F$ , $B D$ and $C E$ .
Hint
Let $A F = A E = x$ , $B F = B D = y$ and $C D = C E = z$ , and form three equations.
Answer
Equal tangents: $A F = A E = x$ , $B F = B D = y$ , $C D = C E = z$ . Then $x + y = 12$ , $y + z = 8$ , $z + x = 10$ . Adding, $2 (x + y + z) = 30$ , so $x + y + z = 15$ . Hence $z = 15 - 12 = 3$ , $x = 15 - 8 = 7$ , $y = 15 - 10 = 5$ . So $A F = 7$ cm, $B D = 5$ cm and $C E = 3$ cm.
Q7 · 3 marks · Short answerHarder board level
$P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ . Prove that $∠ A P B = 2∠ O A B$ .
Hint
Find $∠ P A B$ from the isosceles triangle PAB, then use $∠ O A P = 9 0^{\circ}$ .
Answer
$P A = P B$ , so $△ P A B$ is isosceles and $∠ P A B = ∠ P B A = \frac{18 0 ^{\circ} - ∠ A P B}{2} = 9 0^{\circ} - \frac{1}{2} ∠ A P B$ . Since $∠ O A P = 9 0^{\circ}$ (radius ⟂ tangent), $∠ O A B = 9 0^{\circ} - ∠ P A B = \frac{1}{2} ∠ A P B$ . Hence $∠ A P B = 2∠ O A B$ .
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): If $P A$ and $P B$ are tangents from an external point $P$ to a circle with centre $O$ , then $∠ A O P = ∠ B O P$ .
Reason (R): $△ O A P ≅ △ O B P$ by RHS, since $O A = O B$ , $O P$ is common and $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which congruence gives equal angles at O?
Answer
(a). The RHS congruence in R gives $∠ A O P = ∠ B O P$ by CPCT, so R explains A.
Q9 · 4 marks · Case-basedHOTS / competency
Case study: A circular lid fits snugly inside a frame shaped like a right triangle ABC, touching all three sides. The two sides at the right angle B are $A B = 6$ cm and $B C = 8$ cm.
(i) Find $A C$ . (1 mark)
(ii) Using equal tangent lengths, find the radius r of the lid. (2 marks)
(iii) Check your answer by splitting the triangle into three triangles with the centre O as a common vertex. (1 mark)
Hint
From B the two tangent lengths equal r, because the radii to the two touch points and the two sides form a square.
Answer
(i) $A C = 6^{2} + 8^{2} = 10$ cm. (1)
(ii) Let the lid touch AB, BC, CA at P, Q, R. The radii OP and OQ are perpendicular to AB and BC, and $∠ B = 9 0^{\circ}$ , so OPBQ is a square and $B P = B Q = r$ . Then $A P = A R = 6 - r$ and $C Q = C R = 8 - r$ , so $A C = (6 - r) + (8 - r) = 10$ , giving $r = 2$ cm. (2)
(iii) Area of $△ A B C = \frac{1}{2} \times 6 \times 8 = 24$ cm $^{2}$ . Also area $= \frac{1}{2} r (A B + B C + C A) = \frac{1}{2} \times 2 \times 24 = 24$ cm $^{2}$ , which matches. (1)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Theorem 10.2.Show answer
The lengths of tangents drawn from an external point to a circle are equal.
2.Which congruence rule proves it?Show answer
RHS.
3.What does $O P$ do to $∠ A P B$ ?Show answer
It bisects it.
4.Tangent length formula from $P$ ?Show answer
$O P^{2} - r^{2}$
5.Perimeter of triangle cut off by a third tangent?Show answer
$2 P A$
6.Kind of triangle $P A B$ ?Show answer
Isosceles, as $P A = P B$ .

Lengths of Tangents from an External Point: frequently asked questions

Can I prove Theorem 10.2 using Pythagoras instead of congruence?

Yes, it is a valid proof: both tangent lengths equal the square root of OP squared minus r squared. Still, the RHS congruence proof is the standard one and also gives the angle bisector result, so learn it.

Does the theorem work for a point on the circle?

No. From a point on the circle only one tangent can be drawn, so there is nothing to compare. The theorem needs an external point.

Are constructions of tangents in the 2026-27 syllabus?

No, constructions of tangents are not in the 2026-27 syllabus.

Do I need to draw the figure in the proof?

A neat labelled figure helps the examiner follow you and is expected. Mark the points of contact, the centre and the joined radii.