NCERT Class 9 Maths Solutions: Circles

Question:

Congruent Triangles: Triangles are congruent if all three corresponding sides and all three corresponding angles are equal. Key congruence criteria include SSS, SAS, ASA, AAS.
Properties of Equal Chords: Equal chords are equidistant from the center of the circle.
Angle Properties: Angles subtended by equal chords at the center are equal.


Let the circle have center O. Let AB and CD be two equal chords intersecting at point P within the circle. We need to prove that the line segment OP makes equal angles with the chords AB and CD. This means we need to prove that angle OPA = angle OPC (or angle OPB = angle OPD).

To prove this, we can draw perpendiculars from the center O to the chords AB and CD. Let these perpendiculars be OM and ON respectively, where M is on AB and N is on CD.

Since AB and CD are equal chords, they are equidistant from the center. Therefore, OM = ON.

Now consider the triangles OMP and ONP.
1. OM = ON (as established above, equal chords are equidistant from the center).
2. OP is common to both triangles (OP = OP).
3. Angle OMP = Angle ONP = 90 degrees (since OM is perpendicular to AB and ON is perpendicular to CD).

By the RHS (Right angle-Hypotenuse-Side) congruence criterion, triangle OMP is congruent to triangle ONP.

Since the triangles are congruent, their corresponding angles are equal.
Therefore, angle OPM = angle OPN.

Let’s express these angles in terms of the angles with the chords.
Angle OPM is the angle between the line joining the point of intersection to the center (OP) and the chord AB (or its extension).
Angle OPN is the angle between the line joining the point of intersection to the center (OP) and the chord CD (or its extension).

Thus, the line joining the point of intersection to the centre makes equal angles with the chords.

Question:

Congruent circles: Circles with equal radii.
Congruent chords: Chords of equal length.
Congruent triangles: Triangles with corresponding sides and angles equal.
SSS Congruence Rule: If three sides of one triangle are equal to three corresponding sides of another triangle, then the two triangles are congruent.
Angles at the centre: The angle formed by joining the endpoints of a chord to the centre of the circle.


Let the two congruent circles be Circle 1 and Circle 2, with centres O1 and O2 respectively, and radii r1 and r2.
Since the circles are congruent, their radii are equal, so r1 = r2. Let this common radius be r.

Let AB be a chord in Circle 1 and CD be a chord in Circle 2.
Given that the chords are equal, so AB = CD.

We need to prove that the angles subtended by these equal chords at their respective centres are equal, i.e., ∠AOB = ∠COD.

Consider triangle ΔAOB and triangle ΔCOD.
In ΔAOB:
OA = OB = r (radii of Circle 1)
AB is the chord.

In ΔCOD:
OC = OD = r (radii of Circle 2)
CD is the chord.

Now, we compare ΔAOB and ΔCOD:
1. OA = OC (radii of congruent circles are equal)
2. OB = OD (radii of congruent circles are equal)
3. AB = CD (given that the chords are equal)

Since all three sides of ΔAOB are equal to the corresponding three sides of ΔCOD (SSS congruence rule), we can conclude that ΔAOB ≅ ΔCOD.

By the property of congruent triangles, their corresponding angles are equal. Therefore, the angle subtended by chord AB at the centre O1 (which is ∠AOB) is equal to the angle subtended by chord CD at the centre O2 (which is ∠COD).

Hence, ∠AOB = ∠COD.
This proves that equal chords of congruent circles subtend equal angles at their centres.

Question:

Angles subtended by an arc at the center and 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.


Since ABC and ADC are right triangles with common hypotenuse AC, this implies that points A, B, C, and D all lie on a circle.
The hypotenuse AC of a right triangle inscribed in a circle is always the diameter of the circle.
Therefore, AC is the diameter of the circle passing through points A, B, C, and D.

Now consider the arc ABC. The angle subtended by arc AC at point B is ∠ABC, which is given as 90 degrees.
Similarly, consider the arc ADC. The angle subtended by arc AC at point D is ∠ADC, which is given as 90 degrees.

We need to prove that ∠CAD = ∠CBD.

Let’s consider the arc CD.
The angle subtended by arc CD at point A is ∠CAD.
The angle subtended by arc CD at point B is ∠CBD.

According to the theorem: Angles in the same segment of a circle are equal.
Here, ∠CAD and ∠CBD are angles subtended by the same arc CD in the same segment of the circle.
Therefore, ∠CAD = ∠CBD.

Alternatively, we can use the property of cyclic quadrilaterals.
Since A, B, C, and D lie on a circle, ABCD is a cyclic quadrilateral.
In a cyclic quadrilateral, the sum of opposite angles is 180 degrees.
∠ABC + ∠ADC = 90° + 90° = 180°. This confirms it’s a cyclic quadrilateral.

To prove ∠CAD = ∠CBD, we can consider the arc CD again.
The angle subtended by arc CD at the circumference is ∠CAD and ∠CBD.
Since these angles are subtended by the same arc at the circumference, they are equal.
Hence, ∠CAD = ∠CBD.

Question:

Angle in a semicircle is a right angle. Properties of cyclic quadrilaterals.


Let the triangle be ABC. Let the sides AB and AC be taken as diameters of two circles. Let the circle with diameter AB be C1 and the circle with diameter AC be C2. Let P be the point of intersection of C1 and C2.

Consider the circle C1 with diameter AB. Since P lies on C1, the angle APB is subtended by the diameter AB at a point on the circumference. Therefore, angle APB is a right angle (90 degrees). This means AP is perpendicular to BP.

Similarly, consider the circle C2 with diameter AC. Since P lies on C2, the angle APC is subtended by the diameter AC at a point on the circumference. Therefore, angle APC is a right angle (90 degrees). This means AP is perpendicular to CP.

Now, we have that AP is perpendicular to BP and AP is perpendicular to CP. This implies that B, P, and C lie on a straight line, because if they did not, then from point P, we would have two distinct perpendiculars to the line AP, which is impossible. Therefore, P lies on the line BC.

Alternatively, we can consider the points A, P, B, and C.
In circle C1, angle APB = 90 degrees.
In circle C2, angle APC = 90 degrees.
This means angle APB + angle APC = 90 + 90 = 180 degrees.
Since angle APB and angle APC are adjacent angles sharing a common ray AP, and their sum is 180 degrees, the angles BPC forms a straight line. Therefore, B, P, and C are collinear, which means P lies on the third side BC.

Thus, the point of intersection of these circles lies on the third side.

Question:

Congruent circles have equal radii.
Congruent triangles have equal corresponding sides and angles (SSS, SAS, ASA, AAS congruence criteria).
The area of a triangle can be calculated using two sides and the included angle: Area = 1/2 * a * b * sin(C).


Let the two congruent circles be C1 and C2 with centers O1 and O2 respectively.
Since the circles are congruent, their radii are equal. Let the radius of both circles be ‘r’.
Let AB be a chord in circle C1 and CD be a chord in circle C2.
We are given that chord AB subtends an angle ∠AO1B at the center O1, and chord CD subtends an angle ∠CO2D at the center O2.
We are also given that these angles are equal: ∠AO1B = ∠CO2D.

We need to prove that the chords AB and CD are equal (i.e., AB = CD).

Consider triangle ΔAO1B in circle C1.
The sides O1A and O1B are radii of the circle C1, so O1A = O1B = r.
The angle between these two sides is ∠AO1B.

Consider triangle ΔCO2D in circle C2.
The sides O2C and O2D are radii of the circle C2, so O2C = O2D = r.
The angle between these two sides is ∠CO2D.

Since O1A = O2C (both are radii ‘r’)
And O1B = O2D (both are radii ‘r’)
And ∠AO1B = ∠CO2D (given)

By the Side-Angle-Side (SAS) congruence criterion, triangle ΔAO1B is congruent to triangle ΔCO2D.

Since the triangles are congruent, their corresponding sides are equal.
Therefore, the chord AB in ΔAO1B is equal to the chord CD in ΔCO2D.
AB = CD.

Thus, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Question:

Pythagorean theorem, properties of intersecting circles, geometry of triangles, and the concept of perpendicular bisector.


Let the two circles have centers O1 and O2 and radii r1 = 5 cm and r2 = 3 cm respectively. Let the points of intersection of the two circles be A and B. The common chord is the line segment AB. The distance between the centers is O1O2 = 4 cm.

The line joining the centers of two intersecting circles is the perpendicular bisector of their common chord. Let the common chord AB intersect O1O2 at point M. Therefore, AM = MB, and AB = 2 * AM. Also, angle AMO1 = angle AMO2 = 90 degrees.

Consider triangle AO1O2. The sides of this triangle are AO1 = r1 = 5 cm, AO2 = r2 = 3 cm, and O1O2 = 4 cm. We can check if this is a right-angled triangle using the Pythagorean theorem.
5^2 = 25
3^2 + 4^2 = 9 + 16 = 25
Since 5^2 = 3^2 + 4^2, triangle AO1O2 is a right-angled triangle with the right angle at O2 (angle AO2O1 = 90 degrees).

Now, consider triangle AO1O2 again. AM is the altitude from vertex A to the hypotenuse O1O2.
The area of triangle AO1O2 can be calculated in two ways:
1. (1/2) * base * height = (1/2) * O2O1 * AM = (1/2) * 4 * AM
2. Since angle AO2O1 = 90 degrees, (1/2) * base * height = (1/2) * AO2 * O2O1 = (1/2) * 3 * 4 = 6 sq cm.

Equating the two expressions for the area:
(1/2) * 4 * AM = 6
2 * AM = 6
AM = 3 cm

The length of the common chord AB is 2 * AM.
AB = 2 * 3 cm = 6 cm.

Therefore, the length of the common chord is 6 cm.

Question:

Properties of parallelograms: Opposite angles are equal, opposite sides are parallel and equal, consecutive angles are supplementary.
Properties of cyclic quadrilaterals: Opposite angles are supplementary (sum to 180 degrees).


Let ABCD be a cyclic parallelogram.
Since ABCD is a parallelogram, its opposite angles are equal. Therefore, $\angle A = \angle C$ and $\angle B = \angle D$.
Since ABCD is a cyclic quadrilateral, its opposite angles are supplementary. Therefore, $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$.

Now, substitute the property of parallelogram into the property of cyclic quadrilateral:
Since $\angle A = \angle C$, we can write $\angle A + \angle A = 180^\circ$, which means $2\angle A = 180^\circ$.
Dividing by 2, we get $\angle A = 90^\circ$.
Since $\angle A = 90^\circ$ and $\angle A = \angle C$, then $\angle C = 90^\circ$.

Similarly, since $\angle B = \angle D$, we can write $\angle B + \angle B = 180^\circ$, which means $2\angle B = 180^\circ$.
Dividing by 2, we get $\angle B = 90^\circ$.
Since $\angle B = 90^\circ$ and $\angle B = \angle D$, then $\angle D = 90^\circ$.

Thus, all four angles of the parallelogram ABCD are $90^\circ$. A parallelogram with all angles equal to $90^\circ$ is a rectangle.
Therefore, a cyclic parallelogram is a rectangle.