OP Malhotra Class 9 Maths Solutions Chapter 13 Circle Exercise 13 (A)

 

Question 1. Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.
Answer:

In the given circle, the radius \( OA = 13 \) cm, and the perpendicular distance from the center to the chord \( OM = 12 \) cm.
Let the length of the chord \( AB \) be \( 2x \). Since \( OM \) is perpendicular to \( AB \), \( M \) is the midpoint of \( AB \).
Therefore, \( AM = \frac{AB}{2} = \frac{2x}{2} = x \).
Now, in the right-angled triangle \( \triangle OAM \), we can use the Pythagoras Theorem:
\( OA^2 = OM^2 + AM^2 \)
\( (13)^2 = (12)^2 + x^2 \)
\( 169 = 144 + x^2 \)
\( x^2 = 169 - 144 \)
\( x^2 = 25 \)
\( x = \sqrt{25} \)
\( x = 5 \)
So, the length of chord \( AB = 2x = 2 \times 5 = 10 \) cm. This chord is 10 cm long.
In simple words: We have a right-angled triangle formed by the radius, the distance from the center to the chord, and half the chord. Using the Pythagoras theorem, we found half the chord length. Then we doubled it to get the full chord length.

🎯 Exam Tip: Always remember that a perpendicular from the center of a circle to a chord bisects the chord. This forms a right-angled triangle crucial for applying the Pythagoras theorem.

 

Question 2. In the figure, the radius of the given circle, with centre C, is 6 cm, if the chord AB is 3 cm away from the centre, calculate its length.
Answer:

In the given circle with center C, the radius \( CA = 6 \) cm and the distance of the chord \( AB \) from the center \( CM = 3 \) cm.
Let the length of chord \( AB \) be \( x \) cm. Since \( CM \) is perpendicular to \( AB \), \( M \) is the midpoint of \( AB \).
Therefore, \( AM = \frac{1}{2} AB = \frac{x}{2} \) cm.
Now, in the right-angled triangle \( \triangle CAM \), we use the Pythagoras Theorem:
\( CA^2 = CM^2 + AM^2 \)
\( (6)^2 = (3)^2 + \left(\frac{x}{2}\right)^2 \)
\( 36 = 9 + \frac{x^2}{4} \)
\( \frac{x^2}{4} = 36 - 9 \)
\( \frac{x^2}{4} = 27 \)
\( x^2 = 27 \times 4 \)
\( x^2 = 108 \)
\( x = \sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \)
So, the length of chord \( AB = 6\sqrt{3} \) cm. This demonstrates how chord length changes with its distance from the center.
In simple words: We know the circle's radius and how far the chord is from the center. Using these, and the Pythagoras rule, we find half the chord length. Then we double it to get the total length of the chord.

🎯 Exam Tip: When a chord's distance from the center is given, always connect the center to an endpoint of the chord (radius) and to the chord perpendicularly to form a right-angled triangle for calculations.

 

Question 3. In figure, CD is diameter which meets the chord AB in E, such that AE = BE = 4 cm. If CE is 3 cm, find the radius of the circle.
Answer:

In the given figure, \( CD \) is a diameter of the circle with center \( O \). The chord \( AB \) intersects \( CD \) at point \( E \). We are given that \( AE = BE = 4 \) cm and \( CE = 3 \) cm.
Let \( r \) be the radius of the circle. Then \( OC = r \) and \( OB = r \).
Since \( E \) is on the diameter \( CD \), and \( C \) is a point on the circle, the length \( OE \) can be written as \( OE = OC - CE = r - 3 \).
Now, in the right-angled triangle \( \triangle OEB \), using the Pythagoras Theorem:
\( OB^2 = BE^2 + OE^2 \)
\( r^2 = (4)^2 + (r - 3)^2 \)
\( r^2 = 16 + (r^2 - 6r + 9) \)
\( r^2 = 16 + r^2 - 6r + 9 \)
Now, we subtract \( r^2 \) from both sides and rearrange the terms:
\( r^2 - r^2 + 6r = 16 + 9 \)
\( 6r = 25 \)
\( r = \frac{25}{6} \)
\( r = 4\frac{1}{6} \)
So, the radius of the circle is \( 4\frac{1}{6} \) cm. This calculation shows how the diameter segment and chord length relate to the radius.
In simple words: We set up an equation using the Pythagoras theorem for a triangle inside the circle. We knew the chord's half-length and expressed the distance from the center to the chord in terms of the radius. Solving this equation gave us the radius.

🎯 Exam Tip: When a diameter intersects a chord, the intersection point (E) is often used to define segments for the radius. Clearly define \( OE \) (distance from center to chord) in terms of \( r \) and the given segment lengths.

 

Question 4. In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords, if they are on
(i) the same side of the centre ;
(ii) opposite sides of the centre.
Answer:
Given: Radius of the circle \( R = 5 \) cm.
Length of chord \( AB = 8 \) cm.
Length of chord \( CD = 6 \) cm.
When a perpendicular is drawn from the center to a chord, it bisects the chord.
For chord \( AB \): Half-length \( AL = \frac{1}{2} \times 8 = 4 \) cm.
For chord \( CD \): Half-length \( CM = \frac{1}{2} \times 6 = 3 \) cm.
Let \( OL \) be the distance from center \( O \) to chord \( AB \).
In right-angled \( \triangle OAL \), using Pythagoras Theorem:
\( OA^2 = AL^2 + OL^2 \)
\( (5)^2 = (4)^2 + OL^2 \)
\( 25 = 16 + OL^2 \)
\( OL^2 = 25 - 16 = 9 \)
\( OL = \sqrt{9} = 3 \) cm.
Let \( OM \) be the distance from center \( O \) to chord \( CD \).
In right-angled \( \triangle OCM \), using Pythagoras Theorem:
\( OC^2 = CM^2 + OM^2 \)
\( (5)^2 = (3)^2 + OM^2 \)
\( 25 = 9 + OM^2 \)
\( OM^2 = 25 - 9 = 16 \)
\( OM = \sqrt{16} = 4 \) cm.

(i) When the chords are on the same side of the center:

The distance between the chords \( LM = OM - OL = 4 - 3 = 1 \) cm.

(ii) When the chords are on opposite sides of the center:

The distance between the chords \( LM = OL + OM = 3 + 4 = 7 \) cm. This calculation helps us understand the arrangement of chords within a circle.
In simple words: First, we find the distance of each chord from the center using the radius and half the chord length. Then, if the chords are on the same side, we subtract their distances. If they are on opposite sides, we add their distances to find how far apart they are.

🎯 Exam Tip: Always draw a clear diagram to visualize whether the chords are on the same or opposite sides of the center. This helps correctly apply addition or subtraction to find the distance between them.

 

Question 5. The radius of a circle is 2.5 cm. AB, CF are two parallel chords 3.9 cm apart. If AB = 1.4 cm, find CF.
Answer:

Given the radius of the circle \( R = 2.5 \) cm. Chords \( AB \) and \( CF \) are parallel and 3.9 cm apart. Length of chord \( AB = 1.4 \) cm.
In right-angled \( \triangle OAM \), where \( AM = \frac{AB}{2} = \frac{1.4}{2} = 0.7 \) cm and radius \( OA = 2.5 \) cm.
Using Pythagoras Theorem:
\( OA^2 = AM^2 + OM^2 \)
\( (2.5)^2 = (0.7)^2 + OM^2 \)
\( 6.25 = 0.49 + OM^2 \)
\( OM^2 = 6.25 - 0.49 = 5.76 \)
\( OM = \sqrt{5.76} = 2.4 \) cm.
Since the chords are \( 3.9 \) cm apart and \( OM = 2.4 \) cm, the chords must be on opposite sides of the center for the sum of distances to match. Let \( OL \) be the distance from the center to chord \( CF \).
Then \( OM + OL = 3.9 \)
\( 2.4 + OL = 3.9 \)
\( OL = 3.9 - 2.4 = 1.5 \) cm.
Now, in right-angled \( \triangle OCL \), where radius \( OC = 2.5 \) cm and distance \( OL = 1.5 \) cm.
Using Pythagoras Theorem:
\( OC^2 = CL^2 + OL^2 \)
\( (2.5)^2 = CL^2 + (1.5)^2 \)
\( 6.25 = CL^2 + 2.25 \)
\( CL^2 = 6.25 - 2.25 = 4.00 \)
\( CL = \sqrt{4.00} = 2 \) cm.
The length of chord \( CF = 2 \times CL = 2 \times 2 = 4 \) cm. This demonstrates how finding the distances to chords is essential.
In simple words: We first found how far chord AB is from the center. Since the two chords are 3.9 cm apart, and one is 2.4 cm away, the other chord must be \( 3.9 - 2.4 = 1.5 \) cm away (if they are on opposite sides). Then, using this new distance and the radius, we found half the length of the second chord, and doubled it to get its full length.

🎯 Exam Tip: Always verify if chords are on the same or opposite sides of the center when given their separation. This determines whether you add or subtract the distances from the center to find their relative position.

 

Question 6. A chord distant 2 cm from the centre of a circle is 18 cm long. Calculate the length of a chord of the same circle which is 6 cm distant from the centre.
Answer:

Let the radius of the circle be \( R \).
For the first chord:
Distance from center \( OL = 2 \) cm.
Length of chord \( AB = 18 \) cm, so half-length \( AL = \frac{18}{2} = 9 \) cm.
In right-angled \( \triangle OAL \), using Pythagoras Theorem:
\( OA^2 = AL^2 + OL^2 \)
\( R^2 = (9)^2 + (2)^2 \)
\( R^2 = 81 + 4 = 85 \).
Now, for the second chord:
Distance from center \( OM = 6 \) cm.
Let the half-length of this chord be \( x \) cm. The radius \( R \) is the same for the entire circle.
In a right-angled triangle formed by this chord, its distance from the center, and the radius:
\( R^2 = x^2 + OM^2 \)
\( 85 = x^2 + (6)^2 \)
\( 85 = x^2 + 36 \)
\( x^2 = 85 - 36 \)
\( x^2 = 49 \)
\( x = \sqrt{49} = 7 \) cm.
The length of the second chord is \( 2x = 2 \times 7 = 14 \) cm. This shows how knowing the radius allows us to find other chord properties.
In simple words: First, we used the given chord length and its distance from the center to find the radius of the circle. Then, using this calculated radius and the new distance from the center for the second chord, we found half its length. Finally, we doubled that to get the full length of the second chord.

🎯 Exam Tip: When dealing with multiple chords in the same circle, always find the radius first, as it is a constant for all chords and is usually the key to solving for unknown lengths or distances.

 

Question 7. In figure, circles are concentric with centre O. Find AC.
Answer:

In the figure, two circles share the same center \( O \). The outer circle has radius \( OB = 17 \) cm and the inner circle has radius \( OC = 10 \) cm. The segment \( OM = 9 \) cm is perpendicular to the common chord \( AB \) of the outer circle and chord \( CD \) of the inner circle.
Since \( OM \) is perpendicular to chord \( AB \), \( M \) is the midpoint of \( AB \). In right-angled \( \triangle OMB \):
\( OB^2 = OM^2 + MB^2 \)
\( (17)^2 = (9)^2 + MB^2 \)
\( 289 = 81 + MB^2 \)
\( MB^2 = 289 - 81 = 208 \)
\( MB = \sqrt{208} \approx 14.42 \) cm.
Since \( M \) is the midpoint of \( AB \), \( AM = MB \approx 14.42 \) cm.
Similarly, for the inner circle, \( OM \) is perpendicular to chord \( CD \), making \( M \) the midpoint of \( CD \). In right-angled \( \triangle OMC \):
\( OC^2 = OM^2 + CM^2 \)
\( (10)^2 = (9)^2 + CM^2 \)
\( 100 = 81 + CM^2 \)
\( CM^2 = 100 - 81 = 19 \)
\( CM = \sqrt{19} \approx 4.36 \) cm.
Now, we need to find \( AC \). From the figure, \( AC = AM - CM \).
\( AC \approx 14.42 - 4.36 \)
\( AC \approx 10.06 \) cm. This calculation helps in finding the segment lengths within concentric circles.
In simple words: For the bigger circle, we found half the chord length using its radius and distance from the center. We did the same for the smaller circle. Then, we subtracted the smaller half-chord from the larger one to find the length of segment AC.

🎯 Exam Tip: For concentric circles with a common transversal chord, always use the perpendicular from the center to bisect the chord in both circles. This allows for two separate right-angled triangle calculations to find the necessary segment lengths.

 

Question 8. The length of the common chord of two equal intersecting circles is 10 cm and the distance between the two centres is 6 cm. Find the radius of each circle.
Answer:

Let the two equal circles have centers \( O \) and \( O' \), and let their common chord be \( AB \).
Given: Length of common chord \( AB = 10 \) cm. Distance between centers \( OO' = 6 \) cm.
The line connecting the centers \( OO' \) perpendicularly bisects the common chord \( AB \) at point \( M \).
Therefore, \( AM = MB = \frac{10}{2} = 5 \) cm.
Also, \( M \) is the midpoint of \( OO' \). Since the circles are equal, the line joining their centers is bisected by the common chord.
So, \( OM = MO' = \frac{6}{2} = 3 \) cm.
Now, consider the right-angled triangle \( \triangle OAM \). \( OA \) is the radius of the circle.
Using Pythagoras Theorem:
\( OA^2 = OM^2 + AM^2 \)
\( OA^2 = (3)^2 + (5)^2 \)
\( OA^2 = 9 + 25 \)
\( OA^2 = 34 \)
\( OA = \sqrt{34} \)
\( OA \approx 5.83 \) cm.
Thus, the radius of each circle is approximately \( 5.83 \) cm. This calculation uses the symmetry of intersecting equal circles.
In simple words: When two equal circles cross, their shared chord is cut in half by the line connecting their centers. Also, that line is cut in half by the chord. We used these facts to form a right-angled triangle with half the chord and half the distance between centers, then found the radius using the Pythagoras rule.

🎯 Exam Tip: Remember that for two intersecting equal circles, the line joining their centers perpendicularly bisects their common chord. This creates a powerful right-angled triangle for calculations.

 

Question 9. In figure, AB = 8 cm, CM = 1 cm, CM is the perpendicular bisector of AB. The radius OA = x cm. Find x.
Answer:

In the circle with center \( O \), \( OA \) is the radius, denoted as \( x \) cm. The chord \( AB = 8 \) cm, and \( CM = 1 \) cm. Since \( CM \) is the perpendicular bisector of \( AB \), \( M \) is the midpoint of \( AB \), and \( OM \) is perpendicular to \( AB \).
Therefore, \( AM = \frac{AB}{2} = \frac{8}{2} = 4 \) cm.
Point \( C \) is on the circle and also on the line passing through \( O \) and \( M \). So, \( OC \) is also a radius, \( OC = x \) cm.
From the figure, \( OM = OC - CM = x - 1 \) cm.
Now, in the right-angled triangle \( \triangle OAM \), using Pythagoras Theorem:
\( OA^2 = AM^2 + OM^2 \)
\( x^2 = (4)^2 + (x - 1)^2 \)
\( x^2 = 16 + (x^2 - 2x + 1) \)
\( x^2 = 16 + x^2 - 2x + 1 \)
Subtracting \( x^2 \) from both sides and rearranging:
\( 0 = 16 - 2x + 1 \)
\( 2x = 17 \)
\( x = \frac{17}{2} \)
\( x = 8.5 \)
So, the radius \( x = 8.5 \) cm. This problem connects the radius, chord, and a segment along the perpendicular bisector.
In simple words: We used the given chord length to find its half-length. Then, we expressed the distance from the center to the chord using the radius and the extra segment length. With a right-angled triangle, we applied the Pythagoras rule to find the radius of the circle.

🎯 Exam Tip: Be careful when a point (like C here) lies on the diameter extended from the center to the chord. Express the distance from the center to the chord (OM) correctly in terms of the radius and the given segment (CM).

 

Question 10. In figure, CD is the perpendicular bisector of the chord AB. If AB = 2 cm and CD = 4 cm, calculate the radius of the circle.
Answer:

In the circle with center \( O \), \( AB \) is a chord and \( CD \) is its perpendicular bisector. This means \( M \) is the midpoint of \( AB \), and \( OM \) is perpendicular to \( AB \). The length of chord \( AB = 2 \) cm, so \( AM = \frac{1}{2} \times 2 = 1 \) cm. The length of the segment \( CD = 4 \) cm.
Let \( r \) be the radius of the circle. Then \( OA = r \) and \( OD = r \).
From the figure, if \( CD \) is a segment of the diameter passing through \( M \), and assuming \( M \) is between \( O \) and \( C \), then \( OM = OD - DM \). If \( D \) is one end of diameter, \( C \) is the other.
However, the solution uses `OC = (4-r)`. This `OC` actually represents the distance \( OM \), where `C` is a point such that `OM = 4 - r`.
So, we have \( OM = 4 - r \).
In the right-angled triangle \( \triangle OAM \), using Pythagoras Theorem:
\( OA^2 = OM^2 + AM^2 \)
\( r^2 = (4 - r)^2 + (1)^2 \)
\( r^2 = (16 - 8r + r^2) + 1 \)
\( r^2 = 16 - 8r + r^2 + 1 \)
Subtracting \( r^2 \) from both sides:
\( 0 = 17 - 8r \)
\( 8r = 17 \)
\( r = \frac{17}{8} \)
\( r = 2\frac{1}{8} \) cm.
The radius of the circle is \( 2\frac{1}{8} \) cm. This calculation highlights how careful labeling is important in geometric problems.
In simple words: We know half the chord length. We set up the distance from the center to the chord in terms of the radius and the given segment length. Then, we used the Pythagoras rule in a right-angled triangle to find the exact value of the radius.

🎯 Exam Tip: In diagrams where points on a diameter are given, clarify which segment represents the distance from the center to the chord. Often, a variable is used to express this distance in terms of the radius and other given lengths.

 

Question 11. In figure, AB is chord, OD \( \perp \) AB. BOC is diameter. Prove that CA = 2OD.
Answer:

Given: In the circle, \( AB \) is a chord, and \( OD \perp AB \). \( BOC \) is a diameter, and \( AC \) is joined.
To prove: \( CA = 2OD \).
Proof: Since \( OD \perp AB \), \( D \) is the midpoint of chord \( AB \). (A perpendicular from the center to a chord bisects the chord).
Since \( BOC \) is a diameter, \( O \) is the midpoint of \( BC \).
Now consider \( \triangle ABC \). In this triangle, \( D \) is the midpoint of \( AB \) and \( O \) is the midpoint of \( BC \).
According to the Midpoint Theorem, if a line segment connects the midpoints of two sides of a triangle, it is parallel to the third side and half its length.
Therefore, \( DO \parallel AC \)
\( \implies \) \( DO = \frac{1}{2} AC \)
\( \implies \) \( AC = 2 DO \).
Hence proved. This theorem is useful for relating segments within a triangle inscribed in a circle.
In simple words: We used the Midpoint Theorem. Since D is the middle of AB and O is the middle of BC (because BOC is a diameter), the line DO must be half the length of AC. So, AC is twice the length of DO.

🎯 Exam Tip: The Midpoint Theorem is very useful in circle geometry, especially when a chord is bisected by a perpendicular from the center and a diameter is involved. Look for triangles where two midpoints are connected.

 

Question 12. In figure, OMNP is a square. A circle drawn with centre O cuts the square in X and Y. Prove that : \( \triangle OXM \cong \triangle OYP \). Hence prove that NX = NY.
Answer:

Given: OMNP is a square. A circle with center \( O \) intersects the square at points \( X \) on \( OM \) and \( Y \) on \( OP \).
To prove: (i) \( \triangle OXM \cong \triangle OYP \) and (ii) \( NX = NY \).
Proof:
(i) In right-angled triangles \( \triangle OXM \) and \( \triangle OYP \):
\( OX = OY \) (Both are radii of the same circle)
\( OM = OP \) (Both are sides of the square OMNP)
\( \angle M = \angle P = 90^\circ \) (Angles of a square)
Therefore, by the RHS (Right angle-Hypotenuse-Side) congruence rule:
\( \triangle OXM \cong \triangle OYP \).

(ii) Since \( \triangle OXM \cong \triangle OYP \), their corresponding parts are equal (CPCTC):
\( MX = PY \).
Also, from the properties of a square, \( MN = PN \) (Opposite sides of a square are equal).
Now, consider the segments \( NX \) and \( NY \):
\( NX = MN - MX \)
\( NY = PN - PY \)
Since \( MN = PN \) and \( MX = PY \), it follows that \( MN - MX = PN - PY \).
\( \implies NX = NY \).
Hence proved. This demonstrates the power of congruence in proving other geometric relationships.
In simple words: We first showed that two triangles formed within the square and circle are identical because they have a right angle, matching radii (hypotenuses), and equal square sides. Because they are identical, their remaining sides are equal. Using this, and knowing all sides of a square are equal, we proved that the remaining parts of the square's sides are also equal.

🎯 Exam Tip: When dealing with squares and circles intersecting, look for congruent triangles, often solvable by RHS (Right angle-Hypotenuse-Side) if radii and square sides are involved. CPCTC (Corresponding Parts of Congruent Triangles are Congruent) is key for the second part of such proofs.

 

Question 13. Find the centre of given circle.
Answer:

To find the center of a given circle, follow these steps of construction:
(i) Take three non-collinear points, \( A \), \( B \), and \( C \), on the circumference of the circle.
(ii) Join \( AB \) and \( AC \) to form two chords.
(iii) Draw the perpendicular bisector of chord \( AB \). Each point on this bisector is equidistant from \( A \) and \( B \).
(iv) Draw the perpendicular bisector of chord \( AC \). Each point on this bisector is equidistant from \( A \) and \( C \).
(v) The point where these two perpendicular bisectors intersect is the center of the circle, labeled \( O \).
This works because the perpendicular bisector of any chord of a circle always passes through the center of the circle. This is a fundamental property of circles.
In simple words: To find the middle of a circle, pick three points on its edge. Connect them to make two lines. Draw a line that cuts each of these two lines exactly in half and at a perfect right angle. Where these two lines cross is the center of the circle.

🎯 Exam Tip: Remember that any perpendicular bisector of a chord always passes through the center. So, two such bisectors are enough to pinpoint the center's exact location.

 

Question 14. Draw an arc of a circle to pass through three points A, B and C not in the same straight line.
Answer:

To draw an arc of a circle passing through three non-collinear points \( A \), \( B \), and \( C \), follow these steps:
(i) First, plot the three points \( A \), \( B \), and \( C \) on a plane.
(ii) Join \( AB \) and \( BC \) to form two line segments (chords of the desired circle).
(iii) Draw the perpendicular bisector of \( AB \).
(iv) Draw the perpendicular bisector of \( BC \).
(v) The point where these two perpendicular bisectors intersect is the center of the circle, let's call it \( O \).
(vi) With \( O \) as the center and radius \( OA \) (or \( OB \) or \( OC \), as they will all be equal), draw an arc that passes through points \( A \), \( B \), and \( C \). This arc is the required part of the circle. The center found is equidistant from all three points.
In simple words: First, mark the three points. Connect them to make two lines. Draw lines that cut each of these two lines exactly in half and at a right angle. Where these cutting lines meet is the center of your arc. Then, use that center and the distance to any of the three points as your radius to draw the arc.

🎯 Exam Tip: This construction relies on the property that the center of a circle is equidistant from all points on its circumference. Hence, constructing perpendicular bisectors of chords accurately finds this equidistant center.

 

Question 17. In the given figure, O and O' are the centres of two intersecting circles and APB is parallel to OO'. Prove that AB = 2OO'.
Answer: Given: Two circles with centers O and O' cross each other. A line segment APB is parallel to the line connecting the centers, OO'.
To prove: The length of AB is double the length of OO'.
Construction: From center O, draw a line \( OM \perp AP \). From center O', draw a line \( O'N \perp PB \).
Proof:
Since \( OM \perp AP \), M is the midpoint of AP. So, \( AM = MP = \frac{1}{2}AP \).
Since \( O'N \perp PB \), N is the midpoint of PB. So, \( PN = NB = \frac{1}{2}PB \).
The line segment MN connects the midpoints of AP and PB. Since APB is parallel to OO', and OM and O'N are perpendicular to APB, the figure OMN O' is a rectangle.
Therefore, the distance MN is equal to the distance OO'.
Now, let's look at the length of AB:
\( AB = AP + PB \)
Substitute \( AP = 2 MP \) and \( PB = 2 PN \):
\( AB = 2 MP + 2 PN \)
\( \implies AB = 2 (MP + PN) \)
From the diagram, \( MP + PN \) is the length of MN.
\( \implies AB = 2 MN \)
Since \( MN = OO' \) (from the rectangle OMN O'),
\( \implies AB = 2 OO' \).
This means the length of chord AB is twice the distance between the centers OO'. This proves the statement.
In simple words: When two circles cross, and a line parallel to their center line is drawn, its length is exactly double the distance between the two centers. We show this by using perpendicular lines from the centers and forming a rectangle.

🎯 Exam Tip: Remember that a perpendicular from the center of a circle to a chord always bisects the chord. This is a fundamental property used in many proofs involving circles.

 

Question 18. Two equal chords AB and CD of a circle intersect at P, show that AP = PD and BP = CP.
Answer: Given: In a circle with center O, chords AB and CD are equal (\( AB = CD \)). They intersect each other at point P inside the circle.
To prove: \( AP = PD \) and \( BP = CP \).
Construction: Draw \( OM \perp AB \) and \( ON \perp CD \). Join OP.
Proof:
1. Since \( OM \perp AB \), M is the midpoint of AB. So, \( AM = MB = \frac{1}{2}AB \).
2. Since \( ON \perp CD \), N is the midpoint of CD. So, \( CN = ND = \frac{1}{2}CD \).
3. We are given that \( AB = CD \). Equal chords are always equidistant from the center. Therefore, \( OM = ON \).
4. Consider the right-angled triangles \( \triangle OMP \) and \( \triangle ONP \):
- \( \angle OMP = \angle ONP = 90^\circ \) (By construction)
- OP = OP (This is a common hypotenuse for both triangles)
- \( OM = ON \) (From step 3)

\( \implies \triangle OMP \cong \triangle ONP \) (By RHS (Right angle-Hypotenuse-Side) congruence rule).
5. By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), we get \( PM = PN \).
Now, to prove \( AP = PD \):
We know \( AB = CD \) (given).
Also, \( AP = AB - PB \) and \( PD = CD - PC \).
So, \( AB - PB = CD - PC \).
Since \( AB = CD \), by subtracting equal parts from equal wholes, the remaining parts must be equal.
Thus, \( AP = PD \).
Now, to prove \( BP = CP \):
From step 2, \( MB = \frac{1}{2}AB \) and \( NC = \frac{1}{2}CD \). Since \( AB = CD \), it follows that \( MB = NC \).
We know \( BP = MB - MP \) and \( CP = NC - NP \).
Since \( MB = NC \) and \( MP = NP \) (from step 5), we can say:
\( BP = MB - MP \) and \( CP = MB - MP \).
Thus, \( BP = CP \).
Hence proved.
In simple words: When two chords that have the same length cross inside a circle, the parts they are divided into are equal. The piece of one chord from the start to the crossing point is equal to the piece of the other chord from the crossing point to its end. Also, the remaining pieces of the chords are equal.

🎯 Exam Tip: Remember that for equal chords intersecting, the key steps involve showing that they are equidistant from the center and proving congruence of triangles formed with the intersection point.

 

Question 19. In the figure, shows circle with centre O, with equal chords AB and CD; OE ⊥CD at H and OF ⊥ AB at G. Prove that EH = GF.
Answer: Given: In a circle with center O, chords AB and CD are equal (\( AB = CD \)). A line \( OE \perp CD \) at point H, and a line \( OF \perp AB \) at point G.
To prove: \( EH = GF \).
Proof:
1. We know that \( OE \perp CD \), so H is the midpoint of CD. Also, \( OF \perp AB \), so G is the midpoint of AB.
2. Since chords AB and CD are equal, their distances from the center must also be equal. So, \( OG = OH \).
3. OE and OF are both radii of the same circle, so they must be equal in length. Thus, \( OE = OF \).
4. Now, we subtract the equal lengths from step 2 from the equal lengths in step 3:
\( OF = OE \)
\( \implies OF - OG = OE - OH \)
Looking at the figure:
\( OF - OG \) gives us the segment \( GF \).
\( OE - OH \) gives us the segment \( EH \).

\( \implies GF = EH \)
So, we have proven that \( EH = GF \).
In simple words: If two chords in a circle are the same length, and you draw lines from the center that cut them at right angles, then a small piece of one radius outside the chord will be the same length as a small piece of the other radius outside its chord.

🎯 Exam Tip: Always clearly state which lines are radii and which are distances from the center, as this helps justify equalities like \( OE=OF \) and \( OG=OH \).

 

Question 20. In the figure, C is the centre of the circle. CB bisects the ∠DBE, CD ⊥ PQ and CE ⊥ RS. Prove that PQ = RS.
Answer: Given: In the circle, C is the center. The line CB divides angle \( \angle DBE \) into two equal parts (\( \angle EBC = \angle DBC \)). Also, \( CD \perp PQ \) and \( CE \perp RS \), where chords PQ and RS intersect at point B.
To prove: \( PQ = RS \).
Proof:
Consider the two triangles, \( \triangle EBC \) and \( \triangle DBC \):
1. \( \angle E = \angle D = 90^\circ \) (Because \( CD \perp PQ \) (at D) and \( CE \perp RS \) (at E)).
2. \( \angle EBC = \angle DBC \) (Given that CB bisects \( \angle DBE \)).
3. BC = BC (This side is common to both triangles).

\( \implies \triangle EBC \cong \triangle DBC \) (By AAS (Angle-Angle-Side) congruence rule).
Since the triangles are congruent, their corresponding parts are equal.

\( \implies CE = CD \) (By CPCTC (Corresponding Parts of Congruent Triangles are Congruent)).
Now, CE and CD represent the perpendicular distances from the center C to the chords RS and PQ, respectively.
A key property of circles is that chords which are the same distance from the center must be equal in length.
Since \( CE = CD \), this means the chords RS and PQ must be equal.

\( \implies PQ = RS \)
Thus, we have proven that the length of chord PQ is equal to the length of chord RS.
In simple words: If a line from the center of a circle cuts an angle in half, and perpendicular lines are drawn from the center to two chords, then these two chords must be equal in length. This is because the perpendicular distances from the center to these chords become equal.

🎯 Exam Tip: When proving chord equality, a common strategy is to first prove that their perpendicular distances from the center are equal, often using triangle congruence.

 

Question 21. In the figure, AB and CD are equal chords of a circle whose centre is O. OM ⊥ AB and ON ⊥ CD. Prove that ∠OMN = ∠ONM.
Answer: Given: In a circle with center O, chords AB and CD are equal (\( AB = CD \)). A line \( OM \perp AB \) and another line \( ON \perp CD \). The points M and N are connected by a line segment MN.
To prove: \( \angle OMN = \angle ONM \).
Proof:
1. We are given that chord AB is equal to chord CD.
2. A property of circles states that equal chords are always the same distance from the center. Since OM is the perpendicular distance to AB and ON is the perpendicular distance to CD, it means \( OM = ON \).
3. Now, consider the triangle \( \triangle OMN \). In this triangle, we have just established that two of its sides are equal: \( OM = ON \).
4. In any triangle, if two sides are equal, then the angles opposite to those sides must also be equal. The angle opposite to side ON is \( \angle OMN \), and the angle opposite to side OM is \( \angle ONM \).

\( \implies \angle OMN = \angle ONM \)
Thus, we have proven that angle \( \angle OMN \) is equal to angle \( \angle ONM \).
In simple words: When you have two chords of the same length in a circle, and you draw perpendicular lines from the center to these chords, then the triangle formed by the center and the feet of these perpendiculars will have two equal angles at the base. This is because the sides from the center to these points are equal.

🎯 Exam Tip: This proof relies on two basic circle theorems: equal chords are equidistant from the center, and in an isosceles triangle, angles opposite to equal sides are equal.