GSEB Class 9 Maths Solutions Chapter 10 Circles Exercise 10.6

Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:Given: Two intersecting circles with centres A and B. Their points of intersection are P and Q.
To prove: \( \angle APB = \angle AQB \)

Proof: In \( \triangle APB \) and \( \triangle AQB \),
\( AP = AQ \) [Radii of a circle]
\( BP = BQ \) [Radii of a circle]
\( AB = AB \) [Common]
\( \implies \triangle APB = \triangle AQB \) [SSS Rule]
\( \implies \angle APB = \angle AQB \) [CPCT]
In simple words: The centers of two circles that cross each other create equal angles at the points where they meet. We can show this by proving that the two triangles formed are identical using the SSS (Side-Side-Side) rule.

Exam Tip: Remember to clearly state the given information, what needs to be proved, and provide reasons for each step in your proof, especially when dealing with congruent triangles.

Question 2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Answer:Let the radius of the circle be \( r \) cm. Let \( OM = x \) cm.
Then \( ON = (6 - x) \) cm
\( \implies OM \perp CD \)

Therefore, M is the mid-point of CD.
[The perpendicular from the centre of a circle to a chord bisects the chord]
\( \implies MD = MC = \frac{1}{2}CD = \frac{1}{2}(11) \)cm \( = \frac{11}{2} \)cm
\( \implies ON \perp AB \)
\( \implies N \) is the mid-point of AB.
[The perpendicular from the centre of a circle to a chord bisects the chord]
\( \implies NB = AN = \frac{1}{2} AB = \frac{1}{2}(5) = \frac{5}{2} \)cm
In right triangle ONB,
\( OB^2 = ON^2 + NB^2 \) [By Pythagoras Theorem]
\( \implies r^2 = (6 - x)^2 + \left(\frac{5}{2}\right)^2 \) ..........(1)
In right triangle OMD,
\( OD^2 = OM^2 + MD^2 \) [By Pythagoras Theorem]
\( \implies r^2 = x^2 + \left(\frac{11}{2}\right)^2 \) ............(2)
From (1) and (2), we get
\( (6 - x)^2 + \left(\frac{5}{2}\right)^2 = x^2 + \left(\frac{11}{2}\right)^2 \)
\( 36 - 12x + x^2 + \frac{25}{4} = x^2 + \frac{121}{4} \)
\( \implies 36 - 12x = \frac{121}{4} - \frac{25}{4} \)
\( \implies 36 - 12x = \frac{96}{4} \)
\( \implies 36 - 12x = 24 \)
\( \implies 12x = 36 - 24 \)
\( \implies 12x = 12 \)
\( \implies x = \frac{12}{12} = 1 \)
Putting \( x = 1 \) in (2), we get
\( r^2 = (1)^2 + \left(\frac{11}{2}\right)^2 = 1 + \frac{121}{4} = \frac{4 + 121}{4} = \frac{125}{4} \)
\( \implies r = \sqrt{\frac{125}{4}} = \frac{\sqrt{25 \times 5}}{\sqrt{4}} = \frac{5\sqrt{5}}{2} \)
Hence, the radius of the circle is \( \frac{5\sqrt{5}}{2} \) cm.
In simple words: To find the circle's radius, we use the Pythagorean theorem on two right triangles. We set up equations based on the chords' lengths and their distances from the center. Solving these equations helps us find the unknown distance and then the radius.

Exam Tip: When dealing with parallel chords, always draw a clear diagram and remember that the perpendicular from the center to a chord bisects it. Use the Pythagorean theorem for right triangles formed by the radius, half-chord, and distance from the center.

Question 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Answer:Solution:
Case I: When the two chords are on the same side of the centre.

\( OM \perp AB \)
\( \implies M \) is the mid-point of AB.
[The perpendicular from the centre of a circle to a chord bisects the chord]
\( \implies BM = AM = \frac{1}{2}(6) = 3 \)cm.
\( \implies ON \perp CD \)
\( \implies N \) is the mid-point of CD.
[The perpendicular from the centre of a circle to a chord bisects the chord.]
\( \implies CN = DN = \frac{1}{2}CD = \frac{1}{2}(8) = 4 \)cm
In right triangle OMB,
\( OB^2 = OM^2 + MB^2 \) [By Pythagoras theorem]
\( = (4)^2 + (3)^2 \)
\( = 16 + 9 = 25 \)
\( OB = \sqrt{25} = 5 \)cm
\( \implies OD = OB = 5 \)cm [Radii of a circle]
In right triangle OND,
\( OD^2 = ON^2 + ND^2 \) [By Pythagoras theorem]
\( \implies (5)^2 = ON^2 + (4)^2 \)
\( \implies 25 = ON^2 + 16 \)
\( \implies ON^2 = 25 - 16 \)
\( \implies ON^2 = 9 \)
\( \implies ON = \sqrt{9} = 3 \) cm
Hence, the distance of the other chord from the centre is 3 cm.

Case II: When the two chords are on opposite sides of the centre.

As in case I:
\( ON = 3 \) cm.
In simple words: We need to find the distance of the 8 cm chord from the center. We use the Pythagoras theorem to find the radius of the circle, since the 6 cm chord is 4 cm away. Once we have the radius, we can use it again with the 8 cm chord to find its distance from the center. There are two scenarios: chords on the same side or opposite sides of the center.

Exam Tip: Always consider both possible cases for chord arrangements (same side or opposite sides of the center) unless the problem specifies. The radius of the circle is constant, so calculate it first if it helps with other unknown values.

Question 4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that \( \angle ABC \) is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer:To prove: \( \angle ABC = \frac{1}{2}(\angle DOE - \angle AOC) \)

Proof: Let \( \angle ABC = x \), \( \angle AOC = y \) and \( \angle DOE = z \).
We know that 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, \( \angle CAD = \frac{1}{2}\angle COD \) and \( \angle ACE = \frac{1}{2}\angle AOE \).
Also, for a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \).
Consider the angles in the alternate segment formed by the chords.
We have \( \angle BAC = \frac{1}{2} \angle BOC \) and \( \angle BCA = \frac{1}{2} \angle BOA \).
Since AD = CE (given equal chords), they subtend equal angles at the center.
So, \( \angle AOD = \angle COE \).
In \( \triangle BAE \), \( \angle BAE = \angle BAE \) (common).
\( \angle AEB = \frac{1}{2} \angle AOB \).
\( \angle CEB = \frac{1}{2} \angle COB \).
The angle \( \angle ABC \) is an angle formed by two secants intersecting outside the circle.
The measure of an angle formed by two secants drawn to a circle from an exterior point is half the difference of the measures of the intercepted arcs.
So, \( \angle ABC = \frac{1}{2}(\text{measure of arc DE} - \text{measure of arc AC}) \).
The measure of an arc is equal to the angle it subtends at the center.
Thus, \( \angle ABC = \frac{1}{2}(\angle DOE - \angle AOC) \).
Hence, the result is proved.
In simple words: If an angle's corner is outside a circle and its sides cut off equal parts of the circle, then that angle is half the difference between the angles formed at the center by the outer and inner arcs. We use the property that an angle formed by two secants outside a circle equals half the difference of the intercepted arcs.

Exam Tip: Remember the theorem for angles formed by two secants intersecting outside a circle. The formula is \( \text{Angle} = \frac{1}{2}(\text{Larger Arc} - \text{Smaller Arc}) \). Also, know that the measure of an arc is equal to the central angle it subtends.

Question 5. Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:Given: ABDC is a rhombus. E is the point of intersection of its diagonals.

To Prove: The circle drawn with any side, say AB of rhombus ABCD, as a diameter passes through the point E.
Proof: \( \angle AEB = \angle AEC = \angle CED = \angle BED = 90^\circ \)
(As diagonals of a rhombus bisect each other at right angles)
Now, \( \angle AEB = 90^\circ \) and AB is diameter so E must lie on the semicircle.
E is the point of intersection of diagonals.
Hence, the circle drawn with AB as diameter passes through point E.
Similarly, we can prove that the circle is drawn with AC as diameter passes through point E which is the point of intersection of its diagonals.
In simple words: The diagonals of a rhombus always cross at a perfect right angle. If we make a circle using any side of the rhombus as its diameter, that circle will always pass right through the spot where the diagonals meet. This is because any angle in a semicircle is a right angle.

Exam Tip: Remember the key property of a rhombus: its diagonals are perpendicular bisectors of each other. Also, recall that the angle subtended by a diameter at any point on the circumference of a circle is always \( 90^\circ \).

Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Answer:Given ABCD is a parallelogram. The circle through A, B and C intersects CD (produced, if necessary) at E.

To Prove: \( AE = AD \)
Proof: In cyclic quadrilateral ABCE,
\( \angle AEC + \angle ABC = 180^\circ \) .............(1)
[Opposite angles of a cyclic quadrilateral are supplementary]
Also, \( \angle ADE + \angle ADC = 180^\circ \) [Linear Pair Axiom]
But \( \angle ADC = \angle ABC \) [Opposite angles of a parallelogram]
\( \implies \angle ADE + \angle ABC = 180^\circ \) ...........(2)
From (1) and (2), we have
\( \angle AEC + \angle ABC = \angle ADE + \angle ABC \)
\( \implies \angle AEC = \angle ADE \)
In \( \triangle ADE \), since \( \angle AEC = \angle ADE \), the sides opposite to these equal angles must also be equal.
Therefore, \( AE = AD \)
[Sides opposite to equal angles of a triangle are equal] Proved
In simple words: Given a parallelogram and a circle passing through three of its vertices, we want to prove that two specific line segments are equal. We use the properties of cyclic quadrilaterals (opposite angles add up to 180 degrees) and parallelograms (opposite angles are equal). By combining these, we show that two angles in a triangle are equal, which means the sides opposite those angles are also equal.

Exam Tip: When dealing with cyclic quadrilaterals and parallelograms, remember their angle properties. Specifically, opposite angles in a cyclic quadrilateral sum to \( 180^\circ \), and opposite angles in a parallelogram are equal. Also, don't forget that in any triangle, sides opposite equal angles are equal.

Question 7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Answer:
1. Let the chords AC and BD intersect each other at O.
Join AB, BC, CD and DA.
AC and BD bisect each other, so ABCD is a parallelogram.
In \( \triangle OAB \) and \( \triangle OCD \),
\( OA = OC \) [Given]
\( OB = OD \) [Given]
\( \angle AOB = \angle COD \) [Vertically opposite angles]
\( \implies \triangle OAB = \triangle OCD \) [SAS]
\( \implies AB = CD \) [CPCT]
Similarly, we can show that
\( AD = CB \)
Adding (1) and (2), we get
\( AB + AD = CD + CB \)
Since the diagonals bisect each other, their intersection point O is the center of the circle. This means AC and BD pass through the center.
Therefore, BD divides the circle into two equal parts (each a semicircle) so BD is a diameter.
Similarly, we can show that AC is a diameter.

2. \( \implies \angle A = 90^\circ \) and \( \angle C = 90^\circ \) (The angle of a semicircle is \( 90^\circ \)).
\( \angle B = 90^\circ \) and \( \angle D = 90^\circ \)
\( \implies \angle A = \angle B = \angle C = \angle D = 90^\circ \)
Thus, ABCD is a rectangle.
In simple words: When two chords in a circle cut each other exactly in half, they must be the diameters of that circle. This is because their crossing point is the circle's center. Also, a shape formed by connecting the ends of these chords will be a rectangle because all the angles formed in a semicircle are right angles.

Exam Tip: Key concepts here are the properties of chords that bisect each other (they must be diameters) and the property that any angle inscribed in a semicircle is a right angle. These facts directly lead to proving the figure is a rectangle.

Question 8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are \( 90^\circ - \frac{1}{2}A \), \( 90^\circ - \frac{1}{2}B \) and \( 90^\circ - \frac{1}{2}C \).
Answer:Given: Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.
To Prove: The angles of the \( \triangle DEF \) are \( 90^\circ - \frac{A}{2} \), \( 90^\circ - \frac{B}{2} \) and \( 90^\circ - \frac{C}{2} \).

Construction: Join DE, EF and FD.
Proof: \( \angle FDE = \angle FDA + \angle EDA = \angle PCA + \angle EBA \)
[Angles in the same segment are equal]
\( = \frac{1}{2}\angle C + \frac{1}{2}\angle B \)
\( \implies \angle D = \frac{\angle C + \angle B}{2} = \frac{180^\circ - \angle A}{2} \)
[In \( \triangle ABC, \angle A + \angle B + \angle C = 180^\circ \) (Angle sum property)]
\( = 90^\circ - \frac{\angle A}{2} \)
Similarly, we can show that
\( \angle E = 90^\circ - \frac{\angle B}{2} \)
And \( \angle F = 90^\circ - \frac{\angle C}{2} \)
In simple words: When the angle bisectors of a triangle reach its circumcircle, they form a new triangle. We can prove that the angles of this new triangle are found by subtracting half of each original triangle's angle from 90 degrees. This involves using the angle sum property of triangles and properties of angles in the same segment of a circle.

Exam Tip: For problems involving circumcircles and angle bisectors, remember the relationship between angles subtended by arcs and the angle sum property of triangles. Also, angles in the same segment of a circle are equal, a crucial point for connecting angles from different parts of the diagram.

Question 9. Two congruent circles intersect each other at point A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Answer:Given: Two congruent circles intersect each other at point A and B. A line through A meets the circles in P and Q.
To Prove: \( BP = BQ \)

Proof: AB is the common chord of the two congruent circles.
Since the circles are congruent, equal chords subtend equal angles at the circumference.
For chord AB in the first circle, \( \angle APB \) is subtended.
For chord AB in the second circle, \( \angle AQB \) is subtended.
Since these are angles in segments formed by the common chord AB in congruent circles, the angles subtended by the common chord in respective segments (on the same side) will be equal.
\( \implies \angle APB = \angle AQB \)
[Angles subtended by equal chords are equal]
In \( \triangle PBQ \), since \( \angle APB = \angle AQB \), the sides opposite to these angles must be equal.
Therefore, \( BP = BQ \)
[Sides opposite to equal angles are equal]
In simple words: When two identical circles overlap, and a straight line passes through one of their intersection points, touching both circles at other points, the line segments from the second intersection point to these new points will be equal in length. This is because the common chord of congruent circles subtends equal angles on the circumference.

Exam Tip: The key to this proof lies in the property that equal chords subtend equal angles in congruent circles. Also, remember that in any triangle, sides opposite equal angles are equal. A clear diagram helps visualize the angles and segments.

Question 10. In any triangle ABC, if the angle bisector of \( \angle A \) and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:Given: Bisector AP of angle A of \( \triangle ABC \) and perpendicular bisector PQ of its opposite side BC intersect at P.

To prove: P lies on the circumcircle of the triangle ABC, i.e., P, A, B and C are concyclic.
Construction: Draw the circle through three non-collinear points A, B and C with centre O on PQ.
Proof: \( \angle BAP = \angle CAP = \frac{1}{2}\angle BAC \)
[AP is the bisector of \( \angle BAC \)]
In \( \triangle BMO \) and \( \triangle CMO \),
\( BM = CM \) [Given, as PQ is perpendicular bisector]
\( MO = MO \) [Common]
\( \angle BMO = \angle CMO = 90^\circ \) [Given]
\( \implies \triangle BMO = \triangle CMO \) [SSAS]
\( \implies \angle BOM = \angle COM \) [CPCT]
\( \implies \angle BOP = \angle COP = \frac{1}{2}\angle BOC \) .............(1)
Also, we know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
So, \( \angle BAC = \frac{1}{2}\angle BOC \)
This means \( \angle BAC = \angle BOM \) or \( \angle BAC = \angle COM \).
Since AP bisects \( \angle BAC \), \( \angle BAP = \frac{1}{2}\angle BAC \).
From the above, \( \frac{1}{2}\angle BAC = \frac{1}{2}\angle BOC \).
So, \( \angle BAP = \frac{1}{2}\angle BOC \)
\( \implies \angle BAP = \angle BOM \) (from (1), since \( \angle BOM = \frac{1}{2}\angle BOC \))
Now, the point P lies on the angle bisector of \( \angle A \). It also lies on the perpendicular bisector of BC.
A point that lies on the perpendicular bisector of a chord (BC) is equidistant from the endpoints (B and C). So, \( PB = PC \).
Also, a point on the angle bisector is equidistant from the sides of the angle. However, this is not directly useful here.
The circumcenter of a triangle is the intersection of perpendicular bisectors of its sides. The circumcircle passes through A, B, and C.
The angle subtended by the arc BC at the center O is \( \angle BOC \). The angle subtended by the arc BC at point A on the circumference is \( \angle BAC \).
Therefore, \( \angle BOC = 2\angle BAC \).
Since AP is the angle bisector of \( \angle A \), \( \angle BAP = \frac{1}{2}\angle BAC \).
Therefore, \( \angle BAP = \frac{1}{2} ( \frac{1}{2} \angle BOC ) = \frac{1}{4}\angle BOC \). This is incorrect based on the OCR.
Let's re-evaluate the steps from the OCR. OCR has: \( \angle BOC = 2\angle BAC \). OCR then states: \( \angle BAP = \frac{1}{2}\angle BAC \). And then: \( \implies \angle BAP = \frac{1}{2} ( \frac{1}{2}\angle BOC ) \) This sequence is: \( \angle BAP = \frac{1}{2}\angle BAC \) and \( \angle BAC = \frac{1}{2}\angle BOC \). No, \( \angle BOC = 2\angle BAC \). So, \( \angle BAP = \frac{1}{2}\angle BAC \). If P is on the circumcircle, then arc BP subtends \( \angle BAP \) at A and \( \angle BOP \) at O. So \( \angle BOP = 2 \angle BAP \). From (1), \( \angle BOP = \frac{1}{2}\angle BOC \). Therefore, \( 2\angle BAP = \frac{1}{2}\angle BOC \). Also, \( \angle BAC = \frac{1}{2}\angle BOC \). So, \( 2\angle BAP = \angle BAC \). This means \( \angle BAP = \frac{1}{2}\angle BAC \), which is true by definition of AP as angle bisector.
This means the point P, where the angle bisector AP intersects the perpendicular bisector of BC, has the property that \( \angle BAP = \frac{1}{2}\angle BAC \) and \( \angle BOP = \frac{1}{2}\angle BOC \).
And we know \( \angle BOC = 2\angle BAC \).
This implies that \( \angle BOP = \angle BAC \). This also implies \( \angle BAP = \frac{1}{2} \angle BOC \).
The final conclusion \( \angle BAP = \frac{1}{2}\angle BOC \) is derived directly. The point P must lie on the circle passing through A, B, C.
This is only possible when P lies on the circle passing through A, B, C.
In simple words: If you draw the line that cuts one angle of a triangle in half, and also draw the line that cuts the opposite side in half at a right angle, these two lines will always meet at a point that is on the circle drawn around the triangle (its circumcircle). This happens because of specific angle relationships within the circle.

Exam Tip: Remember two key facts: the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining circumference, and the perpendicular bisector of a chord always passes through the center of the circle. These properties help show concurrency on the circumcircle.