Circle Geometry Practice Problems
This content discusses challenging problems related to coordinate geometry and circles. It covers topics such as proving tangents, finding equations of circles, verifying points on circles, and calculating distances within circles. Detailed solutions and explanations are provided for each problem.
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CHAPTER 15 Coordinate Geometry II: The Circle Solutions: Revision Section B More challenging problems
15 Revision and Exam Style Questions: Section B 1. Prove that the line x 3y + 10 = 0 is a tangent to the circle with equation x2 + y2 = 10 and find the coordinates of the point of contact. Rearrange the linear equation: x 3y + 10 = 0 x = 3y 10 Substitute this linear equation into the quadratic equation: x2 + y2 = 10 (3y 10)2 + y2 = 10 9y2 30y 30y + 100 + y2 = 10 10y2 60y+ 100 10 = 0 ( 10) 10y2 60y + 90 = 0 y2 6y + 9 = 0 y 3 = 0 or y 3 = 0 (y 3)(y 3) = 0 y = 3 y = 3
15 Revision and Exam Style Questions: Section B 1. Prove that the line x 3y + 10 = 0 is a tangent to the circle with equation x2 + y2 = 10 and find the coordinates of the point of contact. Let y = 3: x = 3y 10 x = 3(3) 10 = 9 10 = 1 Point of intersection = ( 1, 3) There is only one point of intersection. Therefore, the line is a tangent.
15 Revision and Exam Style Questions: Section B 2. (i)Show that the points A ( 1, 7) and B (5, 7) lie on a circle whose radius is 5 and centre is (2, 3). Find the equation of the circle: Centre =(2, 3) r = 5 (x h)2 + (y k)2 = r2 (x 2)2 + (y 3)2 = 52 (x 2)2 + (y 3)2 = 25
15 Revision and Exam Style Questions: Section B 2. (i)Show that the points A ( 1, 7) and B (5, 7) lie on a circle whose radius is 5 and centre is (2, 3). Substitute the coordinates of the points A and B into the equation of the circle: A( 1, 7) B(5, 7) (x 2)2 + (y 3)2 = 25 (x 2)2 + (y 3)2 = 25 ( 1 2)2+ (7 3)2 = 25 (5 2)2+ (7 3)2 = 25 ( 3)2 + 42 = 25 32 + 42 = 25 9 + 16 = 25 9 + 16 = 25 25 = 25 25 = 25 In both cases L.H.S. = R.H.S., therefore, the points are on the circle.
15 Revision and Exam Style Questions: Section B 2. (ii) What is the distance from the centre of the circle to the chord [AB]? Draw a sketch of the circle whose radius is 5 and centre is (2, 3) and passes through the points A ( 1, 7) and B (5, 7): From the sketch we can see that the the length of the chord |AB| is 6 units.
15 Revision and Exam Style Questions: Section B 2. (ii) What is the distance from the centre of the circle to the chord [AB]? Alternatively, find the distance between the two points, using the formula: A ( 1, 7) and B (5, 7) (x1, y1) (x2, y2) AB = (x2-x1)2+(y2- y1)2 = (5-(-1))2+(7-7)2 = 62+02 b=1 2AB =1 2(6) =3 = 36 AB =6
15 Revision and Exam Style Questions: Section B 2. (ii) What is the distance from the centre of the circle to the chord [AB]? Use Pythagoras Theorem to find the distance, d, from the centre of the circle to the chord [AB]: d2 + b2 = r2 d2 + 32 = 52 d2 + 9 = 25 d2 = 16 16 d = d = 4 The chord is 4 units from the centre.
15 Revision and Exam Style Questions: Section B 3. sis the circle (x 5)2 + (y + 4)2 = 32. sintersects the x-axis at the points Aand B. Find the coordinates of the points A and B. (i) At the x-axis, y = 0: (x 5)2 + (y + 4)2 = 32 (x 5)2 + (0 + 4)2 = 32 (x 5)2 + 42 = 32 x2 5x 5x + 25 + 16 = 32 x2 10x+ 41 32 = 0 x-intercepts: A: (9, 0) B: (1, 0) x2 10x + 9 = 0 (x 9)(x 1) = 0 x 9 = 0 x = 9 or x 1 = 0 x = 1
15 Revision and Exam Style Questions: Section B 3. sis the circle (x 5)2 + (y + 4)2 = 32. sintersects the x-axis at the points Aand B. Show that |AB| is less than the diameter of s. (ii) A: (9, 0) (x1, y1) (x2, y2) B: (1, 0) AB = (x2-x1)2+(y2- y1)2 r2=32 r = 32 =4 2 = (1-9)2+(0-0)2 Diameter =2(radius) = (-8)2+02 =2(4 2) = 64 =8 2 AB =8 8<8 2 Therefore, AB <diameter
15 Revision and Exam Style Questions: Section B 3. sis the circle (x 5)2 + (y + 4)2 = 32. sintersects the x-axis at the points Aand B. Find the equation of the circle with [AB] as diameter. (iii) Centre of circle = midpoint of [AB] A: (9, 0) (x1, y1) (x2, y2) B: (1, 0) 1 2AB 1 r = x1+x2 2 , y1+ y2 = (8) 2 Midpoint = 2 r = 4 =9+1 , 0+0 2 2 =10 2, 0 2 Centre = (5, 0)
15 Revision and Exam Style Questions: Section B 3. sis the circle (x 5)2 + (y + 4)2 = 32. sintersects the x-axis at the points Aand B. Find the equation of the circle with [AB] as diameter. (iii) Centre = (5, 0), r = 4 Equation: (x h)2 + (y k)2 = r2 (x 5)2 + y2 = 42 (x 5)2 + y2 = 16
15 Revision and Exam Style Questions: Section B 4. The circle, c, has centre ( 1, 2) and radius 5. Write down the equation of c. (i) C: ( 1, 2) r = 5 (x h)2 + (y k)2 = r2 (x ( 1))2 + (y 2)2 = 52 (x + 1)2 + (y 2)2 = 25
15 Revision and Exam Style Questions: Section B 4. (ii)The circle khas equation (x 11)2 + (y 11)2 = 100. Prove that the point A (3, 5) is on cand k. Substitute the coordinates of A(3, 5) into the equation of the circle c: (x h)2 + (y k)2 = r2 (x + 1)2 + (y 2)2 = 25 (3 + 1)2+ (5 2)2 = 25 (4)2 + (3)2 = 25 16 + 9 = 25 25 = 25 Since L.H.S. = R.H.S., the point is on the circle c.
15 Revision and Exam Style Questions: Section B 4. (ii)The circle khas equation (x 11)2 + (y 11)2 = 100. Prove that the point A (3, 5) is on cand k. Substitute the coordinates of A(3, 5) into the equation of the circle k: (x h)2 + (y k)2 = r2 (x 11)2 + (y 11)2 = 100 (3 11)2+ (5 11)2 = 100 ( 8)2+ ( 6)2 = 100 64 + 36 = 100 100 = 100 Since L.H.S. = R.H.S., the point is on the circle k.
15 Revision and Exam Style Questions: Section B 5. cis the circle with centre (0, 0). It passes through the point ( 1, 5), as shown in the diagram. Find the equation of c. (i) C: (0, 0) P: ( 1, 5) (x1, y1) (x2, y2) D = (x2-x1)2+(y2- y1)2 r = (-1-0)2+(5-0)2 (x h)2 + (y k)2 = r2 = (-1)2+52 ( ) 2 26 x2 + y2 = = 1+25 x2 + y2 = 26 r = 26
15 Revision and Exam Style Questions: Section B 5. cis the circle with centre (0, 0). It passes through the point ( 1, 5), as shown in the diagram. The point (k, k) lies inside c, where k . Find all possible values of k. (ii) The centre of the circle is (0, 0) and the radius = 26 =5 099 Therefore, any point which is less than 5 units from (0, 0) will be inside the circle. For example, (2, 2), ( 3, 3), etc. So, possible values of k = 0, 1, 2, 3,
15 Revision and Exam Style Questions: Section B 6. The line x 4y = 0 intersects the circle x2 + y2 = 17 at the points Pand Q. Find the coordinates of P and Q. (i) Rearrange the linear equation: x 4y = 0 x = 4y Substitute the linear equation into the quadratic equation: x2 + y2 = 17 (4y)2 + y2 = 17 16y2 + y2 = 17 17y2 = 17 y2 = 1 y = y = 1 1
15 Revision and Exam Style Questions: Section B 6. The line x 4y = 0 intersects the circle x2 + y2 = 17 at the points Pand Q. Find the coordinates of P and Q. (i) y= 1 y = 1 x = 4y x = 4y x= 4( 1) x = 4(1) x= 4 x = 4 P: ( 4, 1) Q: (4, 1)
15 Revision and Exam Style Questions: Section B 6. The line x 4y = 0 intersects the circle x2 + y2 = 17 at the points Pand Q. Show that [PQ] is a diameter of the circle. (ii) If [PQ] is a diameter, then |PQ| = length of the diameter of the circle P: ( 4, 1)Q: (4, 1) (x1, y1) (x2, y2) r2 = 17 r 17 D = (x2-x1)2+(y2- y1)2 diameter = 2r PQ = (4-(-4))2+(1-(-1))2 2 17 = = (8)2+(2)2 = 64+4 |PQ| = length of the diameter = 68 Therefore, [PQ] is a diameter. =2 17
15 Revision and Exam Style Questions: Section B 7. Joshua claims that if the x-coordinate of the centre of a circle is equal to the length of the radius of the circle, then the y-axis is a tangent to the circle. Investigate Joshua s claim by sketching the circle with centre (3, 2) and radius of length 3. (i)
15 Revision and Exam Style Questions: Section B 7. Joshua claims that if the x-coordinate of the centre of a circle is equal to the length of the radius of the circle, then the y-axis is a tangent to the circle. Is Joshua s claim correct? Justify your answer. (ii) Yes, from the diagram we can see that if the x-coordinate of the centre is 3 and the radius is 3 units, then the centre of the circle is 3 units to the right of the y-axis. Therefore, the circle is touching the y-axis and so the y-axis is a tangent.
15 Revision and Exam Style Questions: Section B 8. The diagram shows two circles cand s, of equal radius. The circles touch at the point P(5, 2). The circle chas centre (0, 0). Find the equation of c. (i) Find the radius of the circle: C: (0, 0) (x1, y1) (x2, y2) P(5, 2) D = (x2-x1)2+(y2- y1)2 r = (5-0)2+(2-0)2 (x h)2 + (y k)2 = r2 ( ) 2 = 52+22 29 x2 + y2 = = 25+4 = 29 x2 + y2 = 29
15 Revision and Exam Style Questions: Section B 8. The diagram shows two circles cand s, of equal radius. The circles touch at the point P(5, 2). The circle chas centre (0, 0). Find the equation of s. (ii) To find the centre of the circle s, Go from (0, 0) into the point P (5, 2) and out the same amount on the far side: +5 +2 +5 +2 (0, 0) (5, 2) (10, 4) Equation of the circle: Circle = (10, 4) r = 29 (x h)2 + (y k)2 = r2 ( ) 2 29 (x 10)2 + (y 4)2 = (x 10)2 + (y 4)2 = 29
15 Revision and Exam Style Questions: Section B 8. The diagram shows two circles cand s, of equal radius. The circles touch at the point P(5, 2). The circle chas centre (0, 0). The line t is a tangent to the circles at the point P. Find the equation of t. (ii) Find the slope of the radius going from the centre of the circle (0, 0) to the point of tangency (5, 2) C(0, 0) P(5, 2) (x1, y1) (x2, y2) Slope=y2- y1 radius tangent mradius mtangent= 1 2 5 x2-x1 =2-0 5-0 =2 5 mtangent= 1 -5 mtangent = 2
15 Revision and Exam Style Questions: Section B 8. The diagram shows two circles cand s, of equal radius. The circles touch at the point P(5, 2). The circle chas centre (0, 0). The line t is a tangent to the circles at the point P. Find the equation of t. (ii) -5 Slope = P(5, 2) 2 y y1 = m(x x1) -5 y 2 = (x 5) 2 2(y 2) = 5(x 5) 2y 4 = 5x + 25 t: 5x + 2y 29 = 0
15 Revision and Exam Style Questions: Section B 9. The circle khas equation (x 2)2 + (y 4)2 = 36. Pand Qare the endpoints of a diameter of kand PQis horizontal. Find the coordinates of P and Q. (i) (x 2)2 + (y 4)2 = 36 Centre: (2, 4) r = 6 The end points of the horizontal diameter are 6 units to the left and to the right of the centre. These points will be: P= ( 4, 4)Q = (8, 4)
15 Revision and Exam Style Questions: Section B 9. The circle khas equation (x 2)2 + (y 4)2 = 36. Pand Qare the endpoints of a diameter of kand PQis horizontal. Hence, or otherwise, write down the equations of the two vertical tangents to k. (ii) Sketch the circle and the two vertical tangents: Vertical tangents are: x= 4 or x = 8 or x 8 = 0 x + 4 = 0
15 Revision and Exam Style Questions: Section B 9. The circle khas equation (x 2)2 + (y 4)2 = 36. Pand Qare the endpoints of a diameter of kand PQis horizontal. Another circle also has these two vertical lines as tangents. The centre of this circle is on the x-axis. Find the equation of this circle. (iii) Since the second circle has the same vertical tangents, it has the same radius and the x-coordinate of the centre is the same as the given circle. Since the centre is on the x-axis, the y-coordinate of the centre = 0 Centre = (2, 0) radius = 6 (x h)2 + (y k)2 = r2 (x 2)2 + y2 = 36
15 Revision and Exam Style Questions: Section B 10. The circle khas equation (x 5)2 + (y + 1)2 = 34. Verify that the point A (8, 4) lies on the circle k. (i) Substitute the coordinates of the point A: (8, 4) into the equation of the circle: (x 5)2 + (y + 1)2 = 34 (8 5)2 + (4 + 1)2 = 34 32 + 52 = 34 9 + 25 = 34 34 = 34 Since L.H.S. = R.H.S., the point is on the circle.
15 Revision and Exam Style Questions: Section B 10. The circle khas equation (x 5)2 + (y + 1)2 = 34. Find the equation of t, the tangent to k at the point A. (ii) Find the slope of the radius from the centre (5, 1) to the point A(8, 4) C: (5, 1) (x1, y1) (x2, y2) Slope=y2- y1 A(8, 4) Radius Tangent x2-x1 =4-(-1) 8-5 =4+1 8-5 =5 3 mradius mtangent= 1 5 3 mtangent= 1 -3 mtangent = 5
15 Revision and Exam Style Questions: Section B 10. The circle khas equation (x 5)2 + (y + 1)2 = 34. A second line, s, is a tangent to k at the point B such that t || s. Find the coordinates of the point B. (iii) Go from A into the centre of the circle and out the same amount on the far side: -3 -5 -3 -5 A (8, 4) (5, 1) B (2, 6)
15 Revision and Exam Style Questions: Section B 10. The circle khas equation (x 5)2 + (y + 1)2 = 34. Hence, or otherwise, find the equation of the line s. (iv) Parallel lines has equal slopes Equation of the line with: -3 t || s mt = ms B(2, 6)ms= 5 -3 y y1 = m(x x1) -3 ms= 5 y ( 6) = (x 2) 5 -3 y + 6 = (x 2) 5 5(y+ 6) = 3(x 2) 5y+ 30 = 3x + 6 s: 3x + 5y + 24 = 0
15 Revision and Exam Style Questions: Section B 10. The circle khas equation (x 5)2 + (y + 1)2 = 34. Find the coordinates of the points where the line s crosses the x and y axes. (v) At the x-axis, y = 0 At the y-axis, x = 0: 3x + 5y + 24 = 0 3x + 5y + 24 = 0 3x + 5(0) + 24 = 0 3(0) + 5y + 24 = 0 3x + 24 = 0 5y + 24 = 0 3x= 24 5y= 24 x= 8 y= 4 8 ( 8, 0) (0, 4 8)
15 Revision and Exam Style Questions: Section B 11. A circle khas equation x2 + y2 = 20. The points P (2, 4), Q ( 2, 4) and R (4, 2) are on the circle. Verify that [PQ] is a diameter of the circle. (i) If [PQ] is a diameter, then |PQ| = length of the diameter P (2, 4) (x1, y1) (x2, y2) D = (x2-x1)2+(y2- y1)2 PQ = Q ( 2, 4) r2 = 20 20 =2 5 r = ( ) + 2 2 ( 2 2) (4 4 ) diameter = 2r 2 2 5 ( = 4 5 ) = (-2-2)2+(4+4)2 = = (-4)2+82 = 16+64 = 80 PQ =4 5 |PQ| = length of the diameter Therefore, [PQ] is a diameter.
15 Revision and Exam Style Questions: Section B 11. A circle khas equation x2 + y2 = 20. The points P (2, 4), Q ( 2, 4) and R (4, 2) are on the circle. Draw the circle, showing the points P, Q and R on a coordinated plane. (ii)
15 Revision and Exam Style Questions: Section B 11. A circle khas equation x2 + y2 = 20. The points P (2, 4), Q ( 2, 4) and R (4, 2) are on the circle. Verify that | PRQ| is a right-angle. (iii) Find the slopes of [PR] and [QR]: Q ( 2, 4) R (4, 2) (x1, y1) (x2, y2) Slope=y2- y1 P (2, 4) (x1, y1) (x2, y2) Slope=y2- y1 x2-x1 =-2-(-4) R (4, 2) x2-x1 =-2-4 4-(-2) =-2-4 4+2 =-6 6 4-2 =-2+4 4-2 =2 2 Slope of PR=1 Slope of QR=-1
15 Revision and Exam Style Questions: Section B 11. A circle khas equation x2 + y2 = 20. The points P (2, 4), Q ( 2, 4) and R (4, 2) are on the circle. Verify that | PRQ| is a right-angle. (iii) If the lines are perpendicular, then PRQ is a right-angle: m1 m2= 1 mQR mPR= 1 1 1 = 1 1 = 1 Therefore PRQ is a right-angle.
15 Revision and Exam Style Questions: Section B 11. A circle khas equation x2 + y2 = 20. The points P (2, 4), Q ( 2, 4) and R (4, 2) are on the circle. Find the equation of the image of the circle k under the translation (1, 3) (4, 7). (iv) Translation from (1, 3) (4, 7) means to increase the x-coordinate by 3 and the y-coordinate by 4: +3 +4 (0, 0) (3, 4) Move the centre (0, 0) by this translation: Centre (3, 4) r = 20 (x h)2 + (y k)2 = r2 ( ) 2 20 (x 3)2 + (y 4)2 = (x 3)2 + (y 4)2 = 20
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. Find the speed of the ship, in kilometres per hour. (i) Find distance between the ship and the submarine at the start: C(0, 0) P(12, 5) (x1, y1) (x2, y2) D= (x2-x1)2+(y2- y1)2 = 169 r =13 r = (12-0)2+(5-0)2 = 122+52 = 144+25
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. Find the speed of the ship, in kilometres per hour. (i) When the ship is on the circle x2 + y2 = 64, its distance from the submarine is equal to the radius of the circle: r2 = 64 r = 8 Distance travelled in the two hour period = 13 8 = 5 km
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. Find the speed of the ship, in kilometres per hour. (i) Time taken = 2 h Speed =Distance Time =5 2 =2 5 km/h
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. Find the equation of the circle, which the ship will be on, at 3 pm, if the speed of the ship remains constant. (ii) 1 hour later (3 pm), the ship has travelled an extra 2 5 km. So it will be 2 5km closer to the submarine, at (0, 0). Radius of new circle: r = 8 2 5 = 5 5 km (x h)2 + (y k)2 = r2 x2 + y2= 5 52 x2 + y2= 30 25
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. If the ship continues to move in a straight line at this speed, at what time, to the nearest minute, will the ship pass over the submarine? (iii) Find the time taken for the ship to travel a distance of 5 5 km: 2 hours and 0 2 h Speed =Distance Time 0 2 60 minutes = 12 minutes Time=Distance 2 h 12 mins from 3 pm Speed 5 5 2 5 Time = 3 + 2 h 12 mins = 2 2 hours = Time = 5:12 pm
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. If the ship doubled its speed from 2 pm onwards, at what time, to the nearest minute, will the ship pass over the submarine? (iv) At 2 pm, the ship is a distance of 8 km from the submarine. Time=Distance Speed = 2 2 5 Speed = 5 km/h =8 5 =1 6 hours 1 hour and 0 6 h 0 6 60 minutes = 36 minutes
15 Revision and Exam Style Questions: Section B 12. A stationary submarine is positioned at the point (0, 0) on a coordinated plane. Using its radar, the submarine can detect an approaching ship. At noon, the ship is at the position (12, 5) on the coordinated plane. Two hours later, the ship has moved closer to the submarine and is now found to be on the circle x2 + y2 = 64. If the ship doubled its speed from 2 pm onwards, at what time, to the nearest minute, will the ship pass over the submarine? (iv) 1 h 36 mins from 2 pm Time = 2 + 1 h 36 mins = 3:36 pm
