Chapter 10: Circles - Board Exam Notes

Exam Weightage & Blueprint

Total: ~6-8 Marks

This chapter is critical for Geometry. The board exams heavily test the two main theorems and their applications in proving geometric properties.

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).

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.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): How many tangents can be drawn from a point inside a circle?

Q2 (2M): A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Find PQ.

Q3 (3M): Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Q4 (4M): 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 sides AB and AC.