Circles, Cyclic Quadrilaterals & Tangentsवृत्त, चक्रीय चतुर्भुज एवं वृत्त की स्पर्श रेखाएँ
Mastery of circles, cyclic quadrilaterals, and tangents is essential for scoring top ranks in the Super TET Junior Mathematics exam. This module emphasizes foundational geometric theorems, angle properties, and rapid computation techniques aligned with SCERT UP and NCERT standards.
Mark complete once covered. Progress syncs automatically with the main syllabus hub.
Syllabus Focus & High-Yield Strategy: Mastery of circles, cyclic quadrilaterals, and tangents is essential for scoring top ranks in the Super TET Junior Mathematics exam. This module emphasizes foundational geometric theorems, angle properties, and rapid computation techniques aligned with SCERT UP and NCERT standards.
1. Fundamental Axioms, Definitions & Foundational Properties
CPA — Center Perpendicular Actively bisects chord.2. Formulas, Standard Theorems & Algorithmic Procedures
| Category / Case | Formula / Mathematical Rule | Topper Calculation Shortcut |
|---|---|---|
| Cyclic Quadrilateral | Sum of opposite angles = $180^circ$ ($angle A + angle C = 180^circ$) | Exterior angle equals interior opposite angle directly. |
| Tangent-Radius Theorem | Tangent $perp$ radius at point of contact ($angle OPT = 90^circ$) | Form right triangle $ riangle OPT$, use $OT^2 = OP^2 + PT^2$. |
3. Word Problem Modeling & Practical Applications
4. Mathematical Edge Cases, Sign Conventions & Common Traps
T-90 — Tangent makes 90 degrees with the radius.Comprehensive Mathematical Matrix & Method Comparison
| Parameter / Dimension | Method / Concept A | Method / Concept B | Exam Speed & Accuracy Takeaway |
|---|---|---|---|
| Chord Property | Perpendicular bisects chord | Equal chords equidistant from center | Use perpendicular bisector for numerical length problems. |
| Angle Theorem | Center angle is double circumference angle | Angles in same segment are equal | Use $2 heta$ relation for center-circumference links. |
| Cyclic Quadrilateral | Opposite angles sum to $180^circ$ | Exterior angle = interior opposite angle | Use exterior angle property for fast angle chasing. |
| Tangent Lengths | Tangents from external point are equal | Tangent perpendicular to radius | Form right triangles to compute unknown distances. |
1. If $AB$ is a chord of a circle with center $O$ and radius $10\text{ cm}$, and the length of chord $AB$ is $16\text{ cm}$, what is the perpendicular distance of the chord from the center?
The perpendicular from the center bisects the chord, so half the chord length is $\frac{16}{2} = 8\text{ cm}$. The radius forms the hypotenuse ($10\text{ cm}$) of a right-angled triangle. By Pythagoras theorem, distance $d = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\text{ cm}$. Option 0 is correct.
2. In a cyclic quadrilateral $ABCD$, if $\angle A = 65^\circ$ and $\angle B = 80^\circ$, what is the measure of $\angle D$?
In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Since $\angle B$ and $\angle D$ are opposite angles, $\angle B + \angle D = 180^\circ \implies 80^\circ + \angle D = 180^\circ \implies \angle D = 100^\circ$. Option 0 is correct.
3. From a point $P$, $10\text{ cm}$ away from the center of a circle, a tangent $PT$ of length $8\text{ cm}$ is drawn. What is the radius of the circle?
The radius $r$, tangent $t$, and distance $d$ form a right-angled triangle with hypotenuse $d = 10\text{ cm}$ and base $t = 8\text{ cm}$. By Pythagoras theorem, $r = \sqrt{d^2 - t^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\text{ cm}$. Option 0 is correct.
4. If angle $\angle AOB = 110^\circ$ subtended by an arc $AB$ at the center $O$ of a circle, what is the angle $\angle ACB$ subtended by the same arc at any point $C$ on the remaining circumference?
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Therefore, $\angle ACB = \frac{1}{2} \angle AOB = \frac{110^\circ}{2} = 55^\circ$. Option 0 is correct.
5. Two concentric circles are of radii $5\text{ cm}$ and $3\text{ cm}$. What is the length of the chord of the larger circle which touches the smaller circle?
The radius of the smaller circle ($3\text{ cm}$) acts as the perpendicular distance from the center to the chord of the larger circle. The radius of the larger circle ($5\text{ cm}$) is the hypotenuse. Half the chord length is $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$. Total chord length = $2 \times 4 = 8\text{ cm}$. Option 0 is correct.
6. If quadrilateral $PQRS$ is cyclic and its exterior angle at vertex $S$ measures $85^\circ$, what is the measure of the interior opposite angle $\angle Q$?
By the exterior angle property of a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle. Thus, interior opposite angle $\angle Q = 85^\circ$. Option 0 is correct.
7. If tangents $PA$ and $PB$ from a point $P$ to a circle with center $O$ are inclined to each other at an angle of $70^\circ$, what is the measure of $\angle POA$?
The line joining the external point to the center bisects the angle between the tangents, so $\angle APO = \frac{70^\circ}{2} = 35^\circ$. In right triangle $\triangle PAO$, $\angle PAO = 90^\circ$. Thus, $\angle POA = 90^\circ - 35^\circ = 55^\circ$. Option 0 is correct.
8. What is the maximum number of common tangents that can be drawn to two non-intersecting circles lying outside each other?
Two non-intersecting external circles admit 4 common tangents: 2 direct common tangents and 2 transverse (internal) common tangents. Option 0 is correct.
9. In a circle of radius $13\text{ cm}$, a chord is at a distance of $5\text{ cm}$ from the center. What is the length of the chord?
Using the right-angled triangle formed by radius ($13$), distance ($5$), and half-chord ($x$): $x = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$. Total chord length = $2 \times 12 = 24\text{ cm}$. Option 0 is correct.
10. If $\text{quadrilateral } ABCD$ is cyclic with $\angle A = 3x$ and $\angle C = x$, what is the value of $x$?
Opposite angles of a cyclic quadrilateral sum to $180^\circ$. Therefore, $\angle A + \angle C = 3x + x = 4x = 180^\circ \implies x = \frac{180^\circ}{4} = 45^\circ$. Option 0 is correct.
5-Minute Exam Simulation Mode
Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.
High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)
Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)
Cyclic Quadrilateral Rule
OPPOSITE
O - Opposite, P - Pairs, P - Produce, O - Supplementary, S - Sum, I - Is, T - Total, E - Eighteen-zero ($180^\circ$).
Tangent and Radius Rule
TPR
T - Tangent, P - Perpendicular, R - Radius.
Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)
Negative Marking Trap 1: Confusing diameter and radius when applying chord-perpendicular distance formulas.
Negative Marking Trap 2: Assuming opposite angles of any quadrilateral sum to $180^\circ$ instead of strictly cyclic quadrilaterals.
Negative Marking Trap 3: Mixing up center angle and circumference angle relationship (multiplying instead of dividing).
Negative Marking Trap 4: Forgetting to multiply half-chord length by 2 to get the total chord length.