Is it correct to say: Taking an inscribed angle equal to the arc it intercepts?

Only the CENTRAL angle equals the arc. An inscribed angle is HALF the intercepted arc.

Is it correct to say: Using AB (the chord) instead of EB (the whole secant) in EA × EB = ET²?

Both segments are measured from the outside point E: EB = EA + AB.

Is it correct to say: Saying ALL opposite angles of any quadrilateral are supplementary?

Only for a CYCLIC quadrilateral (all four vertices on one circle). Check this condition before using it.

Is it correct to say: Adding the arcs for two secants that meet OUTSIDE the circle?

Inside: angle = ½(sum of arcs). Outside: angle = ½(difference of arcs).

Is it correct to say: Forgetting that the radius to the point of contact is perpendicular to the tangent?

Join the centre to the point of contact and mark 90°. Tangent-length questions usually need this right angle.

Is it correct to say: Using d = r₁ + r₂ for circles touching internally?

Externally touching: d = r₁ + r₂. Internally touching: d = r₁ − r₂ (larger minus smaller).

📄 Formula sheet (PDF)

Circle

Tangent theorems, touching circles, arcs, inscribed angle theorem, cyclic quadrilateral, tangent-secant angle and the theorems on intersecting chords, secants and tangents — Maharashtra SSC Geometry Ch 3

📐 17 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→Before using any circle theorem, mark on the figure which arc an angle intercepts. A common error is reading the wrong arc.
→Learn the proofs of the inscribed angle theorem and the cyclic quadrilateral theorem — the second follows in two lines from the first.
→Tangent questions: join the centre to the point of contact. The 90° angle there usually turns the problem into a Pythagoras question.
→For chords and secants, measure every segment FROM the point of intersection E: EA × EB = EC × ED. For a secant from outside, EB is the whole secant, not AB.
→Write the theorem's name as the reason in each step ('by tangent segment theorem', 'opposite angles of a cyclic quadrilateral') — reasons carry marks in geometry.

📐 Formulas(17)

Tangent Theorem

O P ⊥ l (l tangent at P)

Symbol	Meaning
$O$	Centre of the circle
$P$	Point of contact of tangent l

Tangent Segment Theorem★ Board fav

P A = P B

Symbol	Meaning
$P A, P B$	Tangent segments from external point P touching the circle at A and B

Length of a Tangent Segment

P A = O P^{2} - r^{2}

Symbol	Meaning
$O P$	Distance of the external point from the centre (cm)
$r$	Radius (cm)

Touching Circles

d = r_{1} + r_{2} (externally), d = r_{1} - r_{2} (internally, r_{1} > r_{2})

Symbol	Meaning
$d$	Distance between the centres (cm)
$r_{1}, r_{2}$	Radii of the two circles (cm)

Measure of an Arc

m (minor arc A B) = ∠ A O B, m (major arc A B) = 36 0^{\circ} - ∠ A O B

Symbol	Meaning
$∠ A O B$	Central angle subtended by the arc

Sum of Measures of Arcs

m (arc A B C) = m (arc A B) + m (arc B C)

Congruent Arcs and Chords

arc A B ≅ arc C D \Leftrightarrow chord A B ≅ chord C D

Inscribed Angle Theorem★ Board fav

∠ A B C = \frac{1}{2} m (arc A X C)

Corollaries of the Inscribed Angle Theorem

∠ A C B = ∠ A D B (same arc), angle in a semicircle = 9 0^{\circ}

Cyclic Quadrilateral Theorem★ Board fav

∠ A + ∠ C = 18 0^{\circ}, ∠ B + ∠ D = 18 0^{\circ}

Exterior Angle of a Cyclic Quadrilateral

∠ C B E = ∠ A D C (E on A B produced)

Tangent-Secant Angle Theorem

∠ A B C = \frac{1}{2} m (arc A X B)

Angle between Chords Intersecting Inside

∠ A E C = \frac{1}{2} [m (arc A C) + m (arc B D)]

Angle between Secants Intersecting Outside

∠ A E C = \frac{1}{2} [m (arc A C) - m (arc B D)]

Theorem of Internal Division of Chords★ Board fav

A E \times E B = C E \times E D

Symbol	Meaning
$E$	Point of intersection of chords AB and CD (inside the circle)

Theorem of External Division of Chords

A E \times B E = C E \times D E

Symbol	Meaning
$E$	Point of intersection of the secants (outside the circle)

Tangent-Secant Segments Theorem★ Board fav

E A \times E B = E T^{2}

Symbol	Meaning
$E T$	Tangent segment from E
$E A, E B$	Distances from E to the near and far points of the secant

✏️ Solved Examples

1Solved Exampleeasy3 steps

A point P is 13 cm from the centre O of a circle of radius 5 cm. PA and PB are tangents from P. Find PA and PB.

By the tangent theorem, ∠OAP = 90°. Apply Pythagoras in ΔOAP.

P A^{2} = O P^{2} - O A^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144

2Solved Exampleboard3 steps

□ABCD is cyclic. If ∠A = (2x + 10)° and ∠C = (3x − 5)°, find x, ∠A and ∠C.

Opposite angles of a cyclic quadrilateral are supplementary.

(2 x + 10) + (3 x - 5) = 180

3Solved Exampleboard3 steps

Chords AB and CD of a circle intersect at E inside the circle. If AE = 6 cm, EB = 4 cm and CE = 3 cm, find ED and the length of chord CD.

By the theorem of internal division of chords:

A E \times E B = C E \times E D

4Solved ExampleHOTS4 steps

From an external point P, a secant meets a circle at A and B (PA = 4 cm, AB = 5 cm) and a tangent touches it at T. A second secant from P meets the circle at C and D with PC = 3 cm. Find PT and CD.

PB is the whole secant from P.

P B = P A + A B = 4 + 5 = 9 cm

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Taking an inscribed angle equal to the arc it intercepts
✓Only the CENTRAL angle equals the arc. An inscribed angle is HALF the intercepted arc.
2
Using AB (the chord) instead of EB (the whole secant) in EA × EB = ET²
✓Both segments are measured from the outside point E: EB = EA + AB.
3
Saying ALL opposite angles of any quadrilateral are supplementary
✓Only for a CYCLIC quadrilateral (all four vertices on one circle). Check this condition before using it.
4
Adding the arcs for two secants that meet OUTSIDE the circle
✓Inside: angle = ½(sum of arcs). Outside: angle = ½(difference of arcs).
5
Forgetting that the radius to the point of contact is perpendicular to the tangent
✓Join the centre to the point of contact and mark 90°. Tangent-length questions usually need this right angle.
6
Using d = r₁ + r₂ for circles touching internally
✓Externally touching: d = r₁ + r₂. Internally touching: d = r₁ − r₂ (larger minus smaller).

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Two circles of radii 7 cm and 4 cm touch each other. Find the distance between their centres if they touch (i) externally (ii) internally.
Q2
An inscribed angle intercepts an arc of 150°. Find the measure of the inscribed angle.
Q3
In cyclic □PQRS, ∠P = 70°. Find ∠R and the exterior angle formed at R when side QR is produced beyond R.
Q4
Chords AB and CD intersect at E inside a circle. If m(arc AC) = 76° and m(arc BD) = 54°, find ∠AEC.
Q5
A tangent at B and a chord BA make an angle whose intercepted arc measures 140°. Find the angle.
Q6
Secants EAB and ECD meet at E outside a circle. If EA = 5 cm, AB = 7 cm and EC = 6 cm, find ED and CD.

📝 Notes

Circle

This chapter is a chain of theorems. Learn them in the textbook's order and each one becomes the reason for the next.

Tangents

The radius to the point of contact is perpendicular to the tangent.
So two tangents from an external point give two congruent right triangles — the tangent segments are equal.
When two circles touch, the point of contact is on the line of centres, so $d = r_1 + r_2$ (external) or $d = r_1 - r_2$ (internal).

Angles and arcs — the master rule

Every angle result in this chapter is a version of "half the intercepted arc":

Vertex of the angle	Measure
At the centre	equal to the arc
On the circle (inscribed, or tangent-secant)	$\tfrac{1}{2}$ arc
Inside the circle (two chords)	$\tfrac{1}{2}$ (sum of the two arcs)
Outside the circle (two secants)	$\tfrac{1}{2}$ (difference of the two arcs)

The cyclic quadrilateral theorem is the inscribed angle theorem applied twice: the two opposite angles intercept arcs that together make the full 360°, so the angles add to 180°.

Lengths — the product rule

All three length theorems say the same thing: from a point E, the product of the distances to the two points where a line meets the circle is fixed.

E inside: $EA \times EB = EC \times ED$
E outside, two secants: $EA \times EB = EC \times ED$
E outside, a tangent: the two points coincide at T, so $EA \times EB = ET^2$

These are proved with similar triangles (AA test), so the Similarity chapter is the foundation here too.

Quick checks

An inscribed angle is always less than 180° and is half its arc — an arc above 360° or an angle above 180° signals a slip.
In a cyclic quadrilateral, an exterior angle equals the interior opposite angle.
Segment lengths in the product rule are always measured from the point of intersection.