Class 10 Circles Notes

Exam Weightage & Blueprint

Total: ~6-8 Marks

This chapter falls under Unit IV: Geometry (15 marks total). As per the latest syllabus, focus is on: Tangent to a circle at point of contact. Proof that tangent is perpendicular to radius, and proof that tangents from an external point are equal.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Tangent-Radius angle ($90^\circ$), Lengths of tangents
Short Answer	2 or 3	Very High	Proof of Theorem 10.2, Concentric circles, Finding lengths (Pythagoras)
Long Answer	4 or 5	Medium	Quadrilaterals circumscribing circles, Complex proofs

⏰ Last 24-Hour Checklist

Theorem 10.1: Radius $\perp$ Tangent.
Theorem 10.2: Lengths of tangents from external point are equal.
No Tangent: From a point inside the circle.

One Tangent: At a point on the circle.
Two Tangents: From a point outside the circle.
Quad Property: $AB+CD = AD+BC$ (for circumscribing quad).

📐 Important Theorems

Theorem 10.1: Tangent-Radius Perpendicularity

Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Implication: In any problem with a tangent, look for the triangle formed by the radius, tangent, and the line joining the center to the external point. It will be a Right-Angled Triangle. Use Pythagoras Theorem: $$ (Hypotenuse)^2 = (Radius)^2 + (Tangent)^2 $$

Theorem 10.2: Tangents from External Point

Statement: The lengths of tangents drawn from an external point to a circle are equal.

Proof Outline (Often asked for 3 Marks):

1. Join $OP, OQ, OR$.

2. In $\Delta OQP$ and $\Delta ORP$:

- $\angle OQP = \angle ORP = 90^\circ$ (Radius $\perp$ Tangent)

- $OQ = OR$ (Radii)

- $OP = OP$ (Common)

3. $\therefore \Delta OQP \cong \Delta ORP$ (RHS Rule).

4. $\Rightarrow PQ = PR$ (CPCT).

Key Deductions from Theorem 10.2

1. Length of Tangent

If $d$ is the distance to the external point and $r$ is the radius, length $l$ is:

$$ l = \sqrt{d^2 - r^2} $$

2. Circumscribing Quadrilateral

If a quadrilateral ABCD circumscribes a circle, sum of opposite sides is equal:

$$ AB + CD = AD + BC $$

Concept: Number of Tangents

Point Position	Number of Tangents	Diagram Representation
Inside Circle	0	Secant (intersects at 2 points)
On the Circle	1	Unique Tangent
Outside Circle	2	Two tangents of equal length

Solved Examples (Board Marking Scheme)

Q1. Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. (3 Marks)

Step 1: Diagram & Given (0.5 Mark)

Let $C_1$ be the larger circle, $C_2$ the smaller circle. Chord $AB$ of $C_1$ touches $C_2$ at $P$. Center $O$.

Step 2: Tangent Property (1 Mark)

Since $AB$ is a tangent to $C_2$ at $P$ and $OP$ is the radius, $OP \perp AB$ (Theorem 10.1).

Step 3: Chord Property (1.5 Marks)

For the larger circle $C_1$, $AB$ is a chord and $OP \perp AB$.

We know that the perpendicular from the center to a chord bisects the chord.

$\therefore AP = BP$. Hence Proved.

Q2. Two tangents TP and TQ are drawn to a circle with center O from an external point T. Prove that $\angle PTQ = 2\angle OPQ$. (3 Marks)

Step 1: Isosceles Triangle Property (1 Mark)

Let $\angle PTQ = \theta$.

Since $TP = TQ$ (Theorem 10.2), $\Delta TPQ$ is isosceles.

$\therefore \angle TPQ = \angle TQP = \frac{1}{2}(180^\circ - \theta) = 90^\circ - \frac{\theta}{2}$.

Step 2: Tangent Property (1 Mark)

We know $\angle OPT = 90^\circ$ (Radius $\perp$ Tangent).

Step 3: Final Calculation (1 Mark)

$\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - (90^\circ - \frac{\theta}{2}) = \frac{\theta}{2}$.

$\therefore \angle OPQ = \frac{1}{2}\angle PTQ \Rightarrow \angle PTQ = 2\angle OPQ$.

Q3. PQ is a chord of length 8 cm of a circle of radius 5 cm. Tangents at P and Q intersect at T. Find length TP. (4 Marks)

Step 1: Geometry Setup (1 Mark)

Join $OT$ intersecting $PQ$ at $R$. $\Delta TPQ$ is isosceles, so $OT \perp PQ$ and bisects it.

$PR = RQ = 4$ cm.

Step 2: Find OR (1 Mark)

In $\Delta PRO$, $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3$ cm.

Step 3: Similarity/Pythagoras (2 Marks)

Let $TP = x$. In $\Delta PRT$, $x^2 = TR^2 + 4^2$. In $\Delta OPT$, $OT^2 = x^2 + 5^2$.

Alternatively, $\Delta PRO \sim \Delta TRP$ (AA Similarity).

$\frac{TP}{PO} = \frac{RP}{RO} \Rightarrow \frac{x}{5} = \frac{4}{3} \Rightarrow x = \frac{20}{3}$ cm.

Previous Year Questions (PYQs)

2023 (1 Mark): 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.
Ans: Radius $\perp$ Tangent. Use Pythagoras: $r = \sqrt{25^2 - 24^2} = \sqrt{49} = 7$ cm.

2020 (4 Marks): Prove that the parallelogram circumscribing a circle is a rhombus.
Hint: Use property $AB+CD=AD+BC$. For parallelogram, $AB=CD$ and $AD=BC$. Thus $2AB=2AD \Rightarrow AB=AD$. Adjacent sides equal implies Rhombus.

2019 (4 Marks): 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.
Ans: In Quad OAPT, $\angle P = \angle T = 90^\circ$. Sum of angles = $360^\circ$. $\Rightarrow \angle AOB + \angle APB = 180^\circ$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Identifying Hypotenuse: In $\Delta OPT$ (where P is point of contact), the hypotenuse is OT (Center to External Point), NOT the tangent or radius.

Tangents Length: Don't forget that tangents are equal ONLY from an external point.

Diagrams: Losing marks for not drawing or labelling diagrams in proof questions.

💡 Scoring Tips

Mark $90^\circ$: Always mark the radius-tangent angle as $90^\circ$ on your diagram first.

Circumscribing Quads: Remember the property: Sum of opposite sides is equal ($AB+CD = AD+BC$).

Step-wise Proofs: Write "Given", "To Prove", "Construction", and "Proof" clearly for 3-4 mark questions.

📝 More Solved Board Questions

Q4. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. 3 Marks

Sol. Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

We know that tangents from an external point are equal.

$AP = AS$ ...(1), $BP = BQ$ ...(2), $CR = CQ$ ...(3), $DR = DS$ ...(4)

Adding (1), (2), (3), and (4):

$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$

$AB + CD = AD + BC$. Hence Proved.

🎯 Board Pattern (2018–2025): Questions involving "quadrilateral circumscribing a circle" or "parallelogram as a rhombus" are extremely common. Always start by writing the four tangent equality equations.

📋 Board Revision Checklist

✅ Tangent at any point is $\perp$ to radius (Theorem 10.1)
✅ Tangents from an external point are equal (Theorem 10.2)
✅ Tangents from external point subtend equal angles at the center
✅ They are equally inclined to the line joining center to the point
✅ Number of tangents: Inside (0), On (1), Outside (2)
✅ In concentric circles, chord of larger circle is bisected by smaller radius
✅ If $AB+CD = AD+BC$, then quad ABCD circumscribes a circle

💡 Exam Tip:
When proving Theorem 10.2, remember to use RHS congruence criteria. Mentioning the specific congruence rule is often worth 0.5 marks.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. How many tangents can a circle have at most?

2. The distance between two parallel tangents of a circle of radius 4 cm is:

3. If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of $80^\circ$, then $\angle POA$ is:

4. A line intersecting a circle in two points is called a:

5. If the angle between two radii of a circle is $110^\circ$, then the angle between the tangents at the ends of the radii is:

Chapter 10: Circles

Overview