krit.club logo

I'm Up and Down, and Round and Round - How Many Circles?

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A circle is the collection of all points in a plane which are at a fixed distance (radius) from a fixed point (center). A chord is a line segment joining any two points on the circle, with the diameter being the longest chord passing through the center.

A circle showing the center O, a radius, and a chord.
•

The perpendicular from the center of a circle to a chord bisects the chord. Conversely, the line joining the center to the midpoint of a chord is perpendicular to the chord.

Circle with perpendicular OM from center O bisecting chord AB at M.
•

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. This implies that angles in the same segment of a circle are equal.

Circle showing angle AOB at the center and angle ACB at the circumference.
•

A quadrilateral is called cyclic if all its four vertices lie on a circle. The sum of either pair of opposite angles of a cyclic quadrilateral is 180∘180^{\circ}.

📐Formulae

d=2rd = 2r

Length of Chord L=2r2−d2\text{Length of Chord } L = 2\sqrt{r^2 - d^2}

∠AOB=2×∠ACB\angle AOB = 2 \times \angle ACB

∠A+∠C=180∘ (For cyclic quadrilateral ABCD)\angle A + \angle C = 180^{\circ} \text{ (For cyclic quadrilateral } ABCD\text{)}

Circumference C=2πr\text{Circumference } C = 2\pi r

Area A=πr2\text{Area } A = \pi r^2

💡Examples

Problem 1:

In a circle, a chord of length 16 cm16 \text{ cm} is at a distance of 6 cm6 \text{ cm} from the center. Find the radius of the circle.

Solution:

Let the chord be AB=16 cmAB = 16 \text{ cm} and OMOM be the perpendicular from center OO to ABAB. Since OM⊥ABOM \perp AB, MM bisects ABAB. Therefore, AM=162=8 cmAM = \frac{16}{2} = 8 \text{ cm}. In right-angled triangle OMAOMA, by Pythagoras theorem: OA2=OM2+AM2OA^2 = OM^2 + AM^2 OA2=62+82OA^2 = 6^2 + 8^2 OA2=36+64OA^2 = 36 + 64 OA2=100OA^2 = 100 OA=100=10 cmOA = \sqrt{100} = 10 \text{ cm}

Explanation:

We use the property that a perpendicular from the center to a chord bisects it, creating a right-angled triangle where the radius is the hypotenuse.

Problem 2:

If ∠ABC\angle ABC is the angle subtended by an arc ACAC at a point BB on the circle and ∠AOC=130∘\angle AOC = 130^{\circ}, where OO is the center, find ∠ABC\angle ABC.

Solution:

According to the theorem, 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. ∠AOC=2∠ABC\angle AOC = 2 \angle ABC 130∘=2∠ABC130^{\circ} = 2 \angle ABC ∠ABC=130∘2\angle ABC = \frac{130^{\circ}}{2} ∠ABC=65∘\angle ABC = 65^{\circ}

Explanation:

This directly applies the central angle theorem relating the angle at the center and the circumference.

Problem 3:

In a cyclic quadrilateral PQRSPQRS, if ∠P=3x−10∘\angle P = 3x - 10^{\circ} and ∠R=2x+40∘\angle R = 2x + 40^{\circ}, find the value of xx and the angles.

Solution:

In a cyclic quadrilateral, the sum of opposite angles is 180∘180^{\circ}. Since PP and RR are opposite vertices: ∠P+∠R=180∘\angle P + \angle R = 180^{\circ} (3x−10)+(2x+40)=180(3x - 10) + (2x + 40) = 180 5x+30=1805x + 30 = 180 5x=1505x = 150 x=30x = 30 Now, ∠P=3(30)−10=80∘\angle P = 3(30) - 10 = 80^{\circ} and ∠R=2(30)+40=100∘\angle R = 2(30) + 40 = 100^{\circ}.

Explanation:

The sum of opposite angles of a cyclic quadrilateral is always 180∘180^{\circ}.

Problem 4:

In the given figure, OO is the center of the circle. If ∠OAB=30∘\angle OAB = 30^{\circ} and ∠OCB=40∘\angle OCB = 40^{\circ}, find the value of ∠AOC\angle AOC.

A circle with center O and points A, B, C on the circumference forming triangles OAB and OCB.

Solution:

  1. In △OAB\triangle OAB, OA=OBOA = OB (radii of the same circle). Therefore, ∠OBA=∠OAB=30∘\angle OBA = \angle OAB = 30^{\circ}.
  2. In △OCB\triangle OCB, OC=OBOC = OB (radii of the same circle). Therefore, ∠OBC=∠OCB=40∘\angle OBC = \angle OCB = 40^{\circ}.
  3. ∠ABC=∠OBA+∠OBC=30∘+40∘=70∘\angle ABC = \angle OBA + \angle OBC = 30^{\circ} + 40^{\circ} = 70^{\circ}.
  4. By the theorem, the angle subtended by an arc at the center is double the angle at the circumference: ∠AOC=2×∠ABC=2×70∘=140∘\angle AOC = 2 \times \angle ABC = 2 \times 70^{\circ} = 140^{\circ}.

Explanation:

We use the property of isosceles triangles formed by radii and the theorem relating center angles to circumference angles.

Problem 5:

In the figure, ABCDABCD is a cyclic quadrilateral in which ACAC and BDBD are its diagonals. If ∠DBC=55∘\angle DBC = 55^{\circ} and ∠BAC=45∘\angle BAC = 45^{\circ}, find ∠BCD\angle BCD.

Cyclic quadrilateral ABCD with diagonals AC and BD.

Solution:

  1. ∠CAD=∠DBC=55∘\angle CAD = \angle DBC = 55^{\circ} (angles in the same segment subtended by arc CDCD).
  2. ∠DAB=∠CAD+∠BAC=55∘+45∘=100∘\angle DAB = \angle CAD + \angle BAC = 55^{\circ} + 45^{\circ} = 100^{\circ}.
  3. Since ABCDABCD is a cyclic quadrilateral, ∠DAB+∠BCD=180∘\angle DAB + \angle BCD = 180^{\circ}.
  4. ∠BCD=180∘−100∘=80∘\angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ}.

Explanation:

This solution applies the 'angles in the same segment' theorem and the supplementary property of opposite angles in a cyclic quadrilateral.