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Quantitative Aptitude

Circles

Trigonometric Ratios & Identities23 / 51Area & Perimeter
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Concepts (2)

Master tangent properties, secant theorems, and cyclic quadrilateral rules for SSC CGL. Focus on `PT² = PA × PB`, `PA × PB = PC × PD`, and opposite angles in cyclic quads summing to 180° for speed and

Core Formulas

  1. Tangent-Radius Property: A tangent to a circle is perpendicular to the radius at the point of contact. This forms a 90° angle.
  2. Length of Tangents from External Point: If two tangents PA and PB are drawn from an external point P to a circle, then their lengths are equal: PA = PB.
  3. Tangent-Secant Theorem (Power of a Point): If a tangent PT and a secant PAB are drawn from an external point P to a circle, then the square of the length of the tangent is equal to the product of the lengths of the secant segment and its external part: PT² = PA × PB.
  4. Secant-Secant Theorem (Power of a Point): If two secants PAB and PCD are drawn from an external point P to a circle, then the product of the lengths of the segments of one secant is equal to the product of the lengths of the segments of the other secant: PA × PB = PC × PD.
  5. Cyclic Quadrilateral Properties: A quadrilateral whose all four vertices lie on a circle is called a cyclic quadrilateral.
    • The sum of opposite angles is 180°: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
    • The exterior angle is equal to the interior opposite angle.
  6. Alternate Segment Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Worked Example 1

Question: From an external point P, a tangent PT is drawn to a circle with center O and radius 5 cm. If the distance PO = 13 cm, find the length of the tangent PT. Also, if a secant PAB passes through the center O, find PA × PB.

Solution:

  1. Find PT: Since the tangent is perpendicular to the radius at the point of contact T, ΔPTO is a right-angled triangle with ∠PTO = 90°. Using Pythagoras theorem: PO² = PT² + OT² 13² = PT² + 5² 169 = PT² + 25 PT² = 169 - 25 = 144 PT = √144 = 12 cm.
  2. Find PA × PB: The secant PAB passes through the center O. So, A and B are points on the circle. The diameter is AB = 2 * radius = 2 * 5 = 10 cm. PA = PO - OA = 13 - 5 = 8 cm. PB = PO + OB = 13 + 5 = 18 cm. PA × PB = 8 × 18 = 144. Self-check: By Tangent-Secant Theorem, PT² = PA × PB. We found PT² = 144 and PA × PB = 144. The results match.

Worked Example 2

Question: In a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)°, and ∠D = (4x - 5)°. Find the values of x and y.

Solution: For a cyclic quadrilateral, the sum of opposite angles is 180°.

  1. For angles A and C: ∠A + ∠C = 180° (2x + 4) + (2y + 10) = 180 2x + 2y + 14 = 180 2x + 2y = 166 x + y = 83 (Equation 1)

  2. For angles B and D: ∠B + ∠D = 180° (y + 3) + (4x - 5) = 180 4x + y - 2 = 180 4x + y = 182 (Equation 2)

  3. Solve the system of equations: Subtract Equation 1 from Equation 2: (4x + y) - (x + y) = 182 - 83 3x = 99 x = 33

  4. Substitute x into Equation 1: 33 + y = 83 y = 83 - 33 y = 50

Therefore, x = 33 and y = 50.

Shortcuts & Tricks

  • Pythagorean Triplets: For tangent-radius problems, immediately recognize common triplets like (3,4,5), (5,12,13), (7,24,25), (8,15,17) to quickly find unknown sides in right-angled triangles (e.g., in Ex 1, (5,12,13) directly gives PT=12).
  • Direct Application: Most SSC CGL questions are direct applications of these theorems. Identify the setup (tangent-secant, two secants, cyclic quad) and apply the formula without overthinking.
  • Alternate Segment Theorem Visualization: The angle between a tangent and a chord is 'looking' at the arc that the chord subtends. The angle in the alternate segment is also 'looking' at the same arc from the circumference. Visualize this to quickly identify equal angles.
  • Cyclic Quad Exterior Angle: If an exterior angle is given, it's immediately equal to the interior opposite angle. Saves calculation steps.

Common Mistakes

  1. Confusing Power of a Point Theorems: Students often mix up PT² = PA × PB with PA × PB = PC × PD. Remember PT² is only for a tangent, while PA × PB is for secants starting from the same external point.
  2. Incorrectly Identifying Segments: In PA × PB, PA is the external segment and PB is the entire secant length from P to the farthest point on the circle. Don't use AB as PB.
  3. Cyclic Quadrilateral Angle Errors: Assuming adjacent angles sum to 180° or that all angles are 90°. Only opposite angles sum to 180°.
  4. Alternate Segment Theorem Misapplication: Not identifying the correct chord or the correct alternate segment. The angle in the alternate segment must be subtended by the same chord at the circumference.

Derivation (brief)

  • Tangent-Secant Theorem (PT² = PA × PB): Consider triangles ΔPTA and ΔPBT.

    • ∠APT = ∠TPB (Common angle)
    • ∠PTA = ∠PBT (By Alternate Segment Theorem, angle between tangent PT and chord AT is equal to angle subtended by chord AT in the alternate segment, which is ∠ABT or ∠PBT).
    • Since two angles are equal, the third angles must also be equal (∠PAT = ∠PTB).
    • Thus, ΔPTA ~ ΔPBT (by AAA similarity).
    • From similarity, PT/PB = PA/PT = TA/BT. Taking the first two ratios, PT/PB = PA/PT, which implies PT² = PA × PB.

  • Cyclic Quadrilateral Opposite Angles (∠A + ∠C = 180°): Let ABCD be a cyclic quadrilateral. 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.

    • ∠ADC is subtended by arc ABC. Let O be the center. Then reflex ∠AOC = 2∠ADC.
    • ∠ABC is subtended by arc ADC. Then ∠AOC = 2∠ABC.
    • Reflex ∠AOC + ∠AOC = 360° (angles around a point).
    • So, 2∠ADC + 2∠ABC = 360°, which simplifies to ∠ADC + ∠ABC = 180°. Similarly for the other pair of opposite angles.

Advanced Examples

Question: A circle with center O has a chord AB of length 16 cm. The tangents at A and B intersect at P. If the radius of the circle is 10 cm, find the length of PA.

Solution:

  1. Draw radius OA and OB. OA = OB = 10 cm. Since PA and PB are tangents from P, PA = PB. Also, OP bisects ∠APB and is perpendicular to AB at M.
  2. In ΔOAM, OA = 10 cm, AM = AB/2 = 16/2 = 8 cm.
  3. Using Pythagoras in ΔOAM: OA² = OM² + AM² => 10² = OM² + 8² => 100 = OM² + 64 => OM² = 36 => OM = 6 cm.
  4. In ΔOAP, ∠OAP = 90° (tangent perpendicular to radius).
  5. We have two similar triangles: ΔOAP ~ ΔAMP (both are right-angled, and share ∠APM/∠APO). Alternatively, use the property that in a right-angled triangle, the altitude to the hypotenuse creates similar triangles. Here, ΔOAP is right-angled at A. AM is not an altitude to hypotenuse OP. Instead, consider ΔOAP and ΔAMP. No, this is incorrect. A more direct approach: In right-angled ΔOAP, we need AP. We know OA=10. We need OP. Consider ΔOAP and ΔAMP. No. Consider ΔOAM and ΔOAP. They are not directly similar. Let's use the property of similar triangles formed by the tangent and radius. In right-angled ΔOAP, AM is perpendicular to OP (if P,M,O are collinear, which they are). No, AM is perpendicular to OP is incorrect. AM is perpendicular to AB. OP is perpendicular to AB. So AM is not perpendicular to OP. Let's use similarity of ΔOAP and ΔAMP. Not correct. Correct approach: In right-angled ΔOAP, OA² = OM * OP. (This is a property of right-angled triangles, altitude to hypotenuse). No, this is for altitude from right angle to hypotenuse. Here, OA is a leg. Let's use similarity of ΔOAP and ΔAMP. In right-angled ΔOAP (at A), and right-angled ΔAMP (at M, because OP is perpendicular to AB). ∠APM is common to both ΔOAP and ΔAMP. So, ΔOAP ~ ΔAMP (by AA similarity). Therefore, OA/AM = OP/AP = AP/PM. From OA/AM = AP/PM, we don't have PM. From OP/AP = AP/PM, we don't have OP or PM. Let's use OA/AM = OP/AP (This is incorrect, it should be OA/AP = AM/PM or OA/OM = AP/AM). The correct similarity is ΔOAP ~ ΔMAP (∠OAP = ∠AMP = 90°, ∠APO is common). So, OA/MA = OP/PA = PA/MP. From OA/MA = OP/PA: 10/8 = OP/PA. So 5/4 = OP/PA (Eq 1). From OP/PA = PA/MP: OP * MP = PA² (Eq 2). We need OP. In ΔOAP, OP² = OA² + PA² = 10² + PA² = 100 + PA². From Eq 1, OP = (5/4)PA. Substitute into OP² = 100 + PA²: ((5/4)PA)² = 100 + PA² 25/16 PA² = 100 + PA² 25/16 PA² - PA² = 100 (25 - 16)/16 PA² = 100 9/16 PA² = 100 PA² = 100 * 16 / 9 = 1600 / 9 PA = √(1600/9) = 40/3 cm.

Variation Types

  • Finding Lengths: Direct application of tangent/secant theorems to find unknown lengths of tangents, secants, or parts of them.
  • Finding Angles: Using cyclic quadrilateral properties, alternate segment theorem, or tangent-radius property to find unknown angles.
  • Combined Problems: Questions that integrate multiple concepts, e.g., finding a length using power of a point, then using that length in a right-angled triangle formed by a radius and tangent.
  • Area/Perimeter Problems: After finding unknown lengths, calculate the area of a triangle or quadrilateral, or the perimeter of a figure.
  • Inscribed/Circumscribed Figures: Problems involving circles inscribed in quadrilaterals (where tangents from a vertex are equal) or quadrilaterals inscribed in circles (cyclic quads).

Time-Saving Methods

  • Draw Clear Diagrams: A well-drawn diagram (even a quick sketch) helps visualize the problem and apply theorems correctly. Label points and known values immediately.
  • Identify Key Theorems: As soon as you read the problem, identify which theorem(s) are applicable (e.g., 'tangent and secant from external point' -> PT² = PA × PB).
  • Look for Right Angles: The tangent-radius property creates a 90° angle, which often leads to Pythagorean theorem or trigonometric ratios. Always look for these right triangles.
  • Use Options: For multiple-choice questions, sometimes you can test the options, especially for angle problems or when values are simple integers.
  • Mental Math for Squares/Products: Practice quick calculations for squares, square roots, and products involved in power of a point theorems.
  • Recognize Special Triangles: Look for 30-60-90 or 45-45-90 triangles, or Pythagorean triplets, to avoid lengthy calculations.
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Master circle chord theorems: a perpendicular from the center bisects the chord. Use the Pythagorean theorem (r² = d² + L²) to quickly solve for radius, chord length, or distance from the center.

Core Formula

Understanding the relationship between the circle's center, its radius, and a chord is fundamental for SSC CGL. The most crucial theorem and its derived formula are:

  1. Perpendicular from Center to Chord: A perpendicular drawn from the center of a circle to a chord bisects the chord. This means it divides the chord into two equal parts.

  2. Pythagorean Relation: This theorem forms a right-angled triangle with the radius, the perpendicular distance from the center to the chord, and half the chord's length. If r is the radius, d is the perpendicular distance from the center to the chord, and L is half the length of the chord, then: r² = d² + L²

  3. Equal Chords: Chords of equal length are equidistant from the center of the circle. Conversely, chords that are equidistant from the center are equal in length.

Worked Example 1

Question: A chord of length 24 cm is drawn in a circle of radius 13 cm. Find the distance of the chord from the center.

Solution:

  1. Identify given values: Chord length = 24 cm, Radius (r) = 13 cm.
  2. Calculate half the chord length (L): L = 24 cm / 2 = 12 cm.
  3. Apply the Pythagorean relation: r² = d² + L² 13² = d² + 12² 169 = d² + 144
  4. Solve for d²: d² = 169 - 144 = 25
  5. Find d: d = √25 = 5 cm. The distance of the chord from the center is 5 cm.

Worked Example 2

Question: A chord is at a distance of 8 cm from the center of a circle. If the radius of the circle is 17 cm, find the length of the chord.

Solution:

  1. Identify given values: Distance from center (d) = 8 cm, Radius (r) = 17 cm.
  2. Apply the Pythagorean relation: r² = d² + L² 17² = 8² + L² 289 = 64 + L²
  3. Solve for L²: L² = 289 - 64 = 225
  4. Find L: L = √225 = 15 cm.
  5. Calculate the full chord length: Chord length = 2 * L = 2 * 15 = 30 cm. The length of the chord is 30 cm.

Shortcuts & Tricks

  • Pythagorean Triplets: Memorize common Pythagorean triplets like (3,4,5), (5,12,13), (6,8,10), (8,15,17), (7,24,25). These frequently appear in circle problems. In Example 1, (5,12,13) is directly applicable. In Example 2, (8,15,17) is used. Recognizing these saves significant calculation time.
  • Visualize: Always draw a quick sketch. It helps to correctly identify the right-angled triangle and its sides (hypotenuse is always the radius).
  • Direct Application: If a line from the center is perpendicular to a chord, immediately assume it bisects the chord. If a line from the center goes to the midpoint of a chord, assume it's perpendicular.

Common Mistakes

  1. Forgetting to double the half-chord: Many students correctly find L (half-chord length) but forget to multiply it by 2 to get the full chord length, especially in time-pressured exams.
  2. Confusing radius with diameter: Always ensure you are using the radius (r) in the r² = d² + L² formula, not the diameter. If diameter is given, divide by 2.
  3. Incorrectly applying Pythagoras: Ensure the radius r is always the hypotenuse in the r² = d² + L² equation. It's r², not d² or L², that stands alone on one side.

Derivation (brief)

Let's briefly understand why a perpendicular from the center bisects the chord. Consider a circle with center O and a chord AB. Draw a line segment OM perpendicular to AB, where M is a point on AB. Now, draw radii OA and OB. We have two triangles, ΔOMA and ΔOMB.

  1. OA = OB (Both are radii of the same circle).
  2. OM = OM (Common side to both triangles).
  3. ∠OMA = ∠OMB = 90° (By construction, OM is perpendicular to AB).

By the RHS (Right angle-Hypotenuse-Side) congruence criterion, ΔOMA ≅ ΔOMB. Since the triangles are congruent, their corresponding parts are equal (CPCTC). Therefore, AM = MB. This proves that the perpendicular from the center O to the chord AB bisects the chord at M.

Advanced Examples

Question: Two parallel chords of lengths 10 cm and 24 cm are on opposite sides of the center of a circle. If the distance between them is 17 cm, find the radius of the circle.

Solution:

  1. Let the radius be r. Let the distances of the chords from the center be d1 and d2 respectively.
  2. For the first chord (length 10 cm): Half-chord L1 = 10/2 = 5 cm. Using Pythagoras: r² = d1² + 5² => r² = d1² + 25.
  3. For the second chord (length 24 cm): Half-chord L2 = 24/2 = 12 cm. Using Pythagoras: r² = d2² + 12² => r² = d2² + 144.
  4. Since the chords are on opposite sides, the total distance between them is d1 + d2 = 17 cm. So, d2 = 17 - d1.
  5. Equate the expressions for r²: d1² + 25 = d2² + 144 d1² + 25 = (17 - d1)² + 144 d1² + 25 = (289 - 34d1 + d1²) + 144 25 = 289 - 34d1 + 144 25 = 433 - 34d1 34d1 = 433 - 25 = 408 d1 = 408 / 34 = 12 cm.
  6. Substitute d1 back into r² = d1² + 25: r² = 12² + 25 = 144 + 25 = 169 r = √169 = 13 cm. The radius of the circle is 13 cm.

Variation Types

  • Parallel Chords: Problems often involve two parallel chords, either on the same side or opposite sides of the center, requiring careful setup of distances.
  • Chord and Tangent: Sometimes, chord properties are combined with tangent properties, e.g., finding the length of a common chord between two intersecting circles or a chord formed by a tangent and a secant.
  • Central Angle: Questions might provide the central angle subtended by a chord and the radius, asking for the chord length. In such cases, the triangle formed by the two radii and the chord can be solved using trigonometry or special triangle properties (e.g., if the central angle is 60°, it's an equilateral triangle).

Time-Saving Methods

  • Draw and Label: Always quickly sketch the scenario. Label the radius, chord, and perpendicular distance. This visual aid drastically reduces errors.
  • Recognize Special Triangles: Beyond Pythagorean triplets, if a central angle is given, look for 30-60-90 or 45-45-90 triangles. For example, if the central angle is 120 degrees, drawing the perpendicular from the center to the chord creates two 30-60-90 triangles.
  • Practice Mental Calculation: Be quick with squares and square roots of common numbers. This is where memorizing triplets truly shines.
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