Applications — Circumscribed Figures and Angle Relations

Chords and Concentric Circles

Concentric circles share the same centre. A classic result:

A chord of the larger of two concentric circles that touches the smaller circle is bisected at the point of contact.

Why? Let the chord $AB$ of the bigger circle touch the smaller circle at $P$. Then $AB$ is a tangent to the smaller circle at $P$, so $OP \perp AB$ (Theorem 10.1). But a perpendicular from the centre to a chord bisects the chord. Hence $AP = BP$.

If the radii are $R$ (large) and $r$ (small), then each half of the chord is $\sqrt{R^2 - r^2}$, so the full chord length is $$AB = 2\sqrt{R^2 - r^2}.$$

[Board Important] For radii 5 cm and 3 cm, the chord $= 2\sqrt{5^2-3^2} = 2\sqrt{16} = 8$ cm.

The Relation $\angle PTQ = 2\,\angle OPQ$

From an external point $T$, tangents $TP$ and $TQ$ touch a circle with centre $O$. Then $$\angle PTQ = 2\,\angle OPQ.$$

Why? Let $\angle PTQ = \theta$. Since $TP = TQ$, triangle $TPQ$ is isosceles, so $\angle TPQ = \tfrac{1}{2}(180^\circ - \theta) = 90^\circ - \tfrac{\theta}{2}$. Also $\angle OPT = 90^\circ$ (radius ⊥ tangent). Hence $$\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \tfrac{\theta}{2}\right) = \tfrac{\theta}{2} = \tfrac{1}{2}\angle PTQ.$$

So $\angle PTQ = 2\,\angle OPQ$. This appears often as a 3-mark proof.

A Quadrilateral Circumscribing a Circle: $AB + CD = AD + BC$

A quadrilateral $ABCD$ is drawn so that all four sides touch a circle (the circle is inscribed in it). Let the circle touch $AB, BC, CD, DA$ at $P, Q, R, S$.

Using equal tangents from each vertex: $$AP = AS,\quad BP = BQ,\quad CR = CQ,\quad DR = DS.$$

Add them up cleverly: $$AB + CD = (AP+PB) + (CR+RD) = (AS+BQ) + (CQ+DS) = (BQ+CQ)+(AS+DS) = BC + AD.$$

Result: In any quadrilateral circumscribing a circle, the sums of opposite sides are equal: $$AB + CD = AD + BC.$$

[Board Important] A parallelogram circumscribing a circle must be a rhombus: in a parallelogram $AB=CD$ and $AD=BC$, and this result forces $2\,AB = 2\,BC$, i.e. all sides equal.

A Triangle Circumscribing a Circle (the Incircle)

When a circle is inscribed in a triangle $ABC$ (its incircle), each vertex sends two equal tangents to the circle. If the incircle touches $BC, CA, AB$ at $D, E, F$, then $$BD = BF,\quad CD = CE,\quad AE = AF.$$

These equal tangent lengths let you find unknown sides. Combined with the area relation Area $= r \times s$ (where $r$ is the inradius and $s$ the semi-perimeter), triangle-incircle problems are fully solvable.

Key Point: From each vertex, the two tangent segments to the incircle are equal — label them and set up equations.

Tangents at the Ends of a Diameter

The tangents drawn at the two ends of a diameter of a circle are parallel.

Why? Let $AB$ be a diameter. The tangent at $A$ is perpendicular to radius $OA$, i.e. to the line $AB$; the tangent at $B$ is perpendicular to $OB$, i.e. to the same line $AB$. Two lines perpendicular to the same line are parallel.

Another standard result: if $XY$ and $X'Y'$ are two parallel tangents and a third tangent $AB$ touches the circle at $C$ meeting them at $A$ and $B$, then $\angle AOB = 90^\circ$ (using the angle-bisector property at $A$ and $B$).

Solved Examples

Example 1: Chord of concentric circles

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller circle.

Solution:

1. The chord touches the smaller circle, so the perpendicular from the centre (radius 3) bisects it.
2. Half-chord $= \sqrt{5^2 - 3^2} = \sqrt{16} = 4$ cm.
3. Full chord $= 2 \times 4 = 8$ cm.

Final Answer: 8 cm.

Takeaway: Chord $= 2\sqrt{R^2 - r^2}$ for concentric radii $R>r$.

Example 2: Quadrilateral circumscribing a circle

A quadrilateral $ABCD$ circumscribes a circle. If $AB = 6$ cm, $BC = 7$ cm and $CD = 4$ cm, find $AD$.

Solution:

1. Opposite sides are equal in sum: $AB + CD = AD + BC$.
2. $6 + 4 = AD + 7 \Rightarrow AD = 10 - 7 = 3$ cm.

Final Answer: $AD = 3$ cm.

Takeaway: $AB + CD = AD + BC$ for any circumscribing quadrilateral.

Example 3: Triangle circumscribing a circle

A triangle $ABC$ is drawn to circumscribe a circle of radius 4 cm, and the point of contact $D$ divides $BC$ into $BD = 8$ cm and $DC = 6$ cm. Find $AB$ and $AC$.

Solution:

1. Equal tangents: $BF = BD = 8$, $CE = CD = 6$, and let $AF = AE = x$.
2. Sides: $BC = 14$, $AB = 8 + x$, $AC = 6 + x$; semi-perimeter $s = 14 + x$.
3. Area $= r s = 4(14 + x)$. Also Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(14+x)\,x\cdot 8\cdot 6}$.
4. Squaring: $16(14+x)^2 = 48x(14+x) \Rightarrow 16(14+x) = 48x \Rightarrow 14 + x = 3x \Rightarrow x = 7$.
5. $AB = 8 + 7 = 15$ cm, $AC = 6 + 7 = 13$ cm.

Final Answer: $AB = 15$ cm, $AC = 13$ cm.

Takeaway: Equal tangents + Area $= rs$ crack incircle problems.

Example 4: Parallelogram circumscribing a circle

Prove that a parallelogram that circumscribes a circle is a rhombus.

Solution:

1. For a circumscribing quadrilateral $ABCD$: $AB + CD = AD + BC$.
2. In a parallelogram $AB = CD$ and $AD = BC$. Substituting: $2AB = 2BC$, so $AB = BC$.
3. Adjacent sides equal in a parallelogram $\Rightarrow$ all sides equal $\Rightarrow$ rhombus.

Final Answer: It must be a rhombus.

Takeaway: Circumscribing + parallelogram forces all sides equal.

Example 5: Tangents at ends of a diameter

Prove that the tangents at the two ends of a diameter of a circle are parallel.

Solution:

1. Let $AB$ be a diameter. The tangent at $A$ is $\perp OA$; the tangent at $B$ is $\perp OB$.
2. But $OA$ and $OB$ lie along the same line $AB$.
3. Both tangents are perpendicular to the same line $AB$, so they are parallel.

Final Answer: The two tangents are parallel.

Takeaway: Perpendiculars to the same line are parallel.