Tangents, Chords, and Angle Theorems in Circles for CBSE Class 10

When chords and tangents intersect in circle geometry, they create interconnected angular relationships that can be deduced with surgical precision using triangle congruence and parallel line theorems. These geometric riders require students to synthesize multiple theorems from Class 9 (such as angle sum properties and parallel transversals) with Class 10 tangent principles.

In CBSE Class 10 Mathematics, Chapter 10 (Circles), two specific angular theorems appear with remarkable frequency.

Important: <u>The angle between two tangents drawn from an external point to a circle is always twice the angle between the chord joining the points of contact and the radius (ngle PTQ = 2ngle OPQ).</u>: the $90^\circ$ parallel tangents rider ($\angle AOB = 90^\circ$) and the tangent-chord angle theorem ($\angle PTQ = 2\angle OPQ$).

What You Will Learn

• The famous NCERT parallel tangents rider: Proving $\angle AOB = 90^\circ$
• The tangent-chord angle relationship: Proving $\angle PTQ = 2\angle OPQ$
• Tangent perpendicularity at the endpoints of a diameter
• Solved CBSE board examination angle-hunting problems
• Common geometric traps and presentation standards

1. Theorem 1: The Parallel Tangents Rider (ngle AOB = 90^\circ)

Problem Statement (NCERT Question 9 & CBSE Favorite)

$XY$ and $X'Y'$ are two parallel tangents to a circle with center $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that: $\mathbf{\angle AOB = 90^\circ}$

    X ------------ P ------------------ A ------------------- Y
                   |                   /                    |                  /                      |                 /                        O ---------------+       C (Point of Contact)
                   |                 \     /
                   |                  \   /
                   |                   \ /
    X' ----------- Q ------------------ B ------------------- Y'

Step-by-Step Proof:

1. Construction:

Join center $O$ to the point of contact $C$ (draw $OC$). Let the diameter connecting the parallel tangents be $PQ$ (passes through center $O$).

2. Prove Congruence of Top Triangles ($\Delta OPA$ and $\Delta OCA$):

In $\Delta OPA$ and $\Delta OCA$:

• $OP = OC$ (Radii of the same circle)
• $AP = AC$ (Lengths of tangents from external point $A$, Theorem 10.2)
• $OA = OA$ (Common side) Therefore, by the SSS Congruence Criterion: $\Delta OPA \cong \Delta OCA$ By CPCT (Corresponding Parts of Congruent Triangles): $\angle POA = \angle COA \implies \mathbf{\angle PAC = 2\angle COA} \quad \text{--- (1)}$

3. Prove Congruence of Bottom Triangles ($\Delta OQB$ and $\Delta OCB$):

Similarly, in $\Delta OQB$ and $\Delta OCB$:

• $OQ = OC$ (Radii)
• $BQ = BC$ (Tangents from external point $B$)
• $OB = OB$ (Common side) By the SSS Congruence Criterion: $\Delta OQB \cong \Delta OCB$ By CPCT: $\angle QOB = \angle COB \implies \mathbf{\angle QBC = 2\angle COB} \quad \text{--- (2)}$

4. Use the Straight Line Angle Property of Diameter $POQ$:

Notice that $POQ$ is a straight diameter line. Therefore, all angles on this line sum to $180^\circ$: $\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$

Substitute $\angle POA = \angle COA$ and $\angle QOB = \angle COB$: $\angle COA + \angle COA + \angle COB + \angle COB = 180^\circ$ $2\angle COA + 2\angle COB = 180^\circ$ $2(\angle COA + \angle COB) = 180^\circ$

5. Conclude the Proof:

Divide both sides by 2: $\angle COA + \angle COB = \frac{180^\circ}{2} = 90^\circ$ Notice from the diagram that $(\angle COA + \angle COB) = \mathbf{\angle AOB}$. Therefore: $\mathbf{\angle AOB = 90^\circ}$ Hence, proved.

2. Theorem 2: The Tangent-Chord Angle Theorem (ngle PTQ = 2ngle OPQ)

Problem Statement (NCERT Example 2 & CBSE Classic)

Two tangents $TP$ and $TQ$ are drawn to a circle with center $O$ from an external point $T$. Prove that: $\mathbf{\angle PTQ = 2\angle OPQ}$

                                      P
                                     /|                                    / |                              O ----+  |                                      \ |   \ T (External Point)
                                     \|  /
                                      Q /

Step-by-Step Proof:

1. Analyze the Isosceles Triangle $\Delta TPQ$:

• Let $\angle PTQ = \theta$.
• By Theorem 10.2, the lengths of tangents from an external point are equal: $TP = TQ$
• Since $TP = TQ$, $\Delta TPQ$ is an isosceles triangle.
• Therefore, the base angles opposite to these sides are equal: $\angle TPQ = \angle TQP$

2. Apply Angle Sum Property in $\Delta TPQ$:

$\angle PTQ + \angle TPQ + \angle TQP = 180^\circ$ $\theta + 2\angle TPQ = 180^\circ$ $2\angle TPQ = 180^\circ - \theta \implies \mathbf{\angle TPQ = 90^\circ - \frac{\theta}{2}} \quad \text{--- (1)}$

3. Use Radius-Tangent Perpendicularity (Theorem 10.1):

Radius $OP$ is perpendicular to tangent $TP$: $\angle OPT = 90^\circ$ Notice from the diagram that $\angle OPT = \angle OPQ + \angle TPQ$: $\angle OPQ + \angle TPQ = 90^\circ$ $\mathbf{\angle OPQ = 90^\circ - \angle TPQ} \quad \text{--- (2)}$

4. Substitute Equation (1) into Equation (2):

$\angle OPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right)$ $\angle OPQ = 90^\circ - 90^\circ + \frac{\theta}{2} = \frac{\theta}{2}$ $2\angle OPQ = \theta$

5. Replace $heta$ with ngle PTQ:

$\mathbf{\angle PTQ = 2\angle OPQ}$ Hence, proved.

3. Summary and Examination Tips

Exam Tip: In questions involving tangents from an external point $T$ and chord $PQ$, always remember that $\Delta TPQ$ is an isosceles triangle ($TP = TQ$)! Recognizing this immediately unlocks base angle equalities.

Common Mistake: Confusing $\angle OPQ$ with $\angle OPT$. $\angle OPT$ is the FULL $90^\circ$ angle between the radius and the tangent, whereas $\angle OPQ$ is the smaller interior angle between the radius and the chord $PQ$!