HKDSE MATHEMATICS RONALD HUI TAK SUN SECONDARY SCHOOL

HOMEWORK  SHW6-01, 6-A1  All done! Good job!  SHW6-B1, 6-C1  Sam L  SHW6-R1  Sam L, Walter (RD)  SHW6-P1  Tashi, Daniel, Sam L 22 October 2015 Ronald HUI

HOMEWORK  RE6  Tashi, Vincent, Daniel, Matthew, Kelvin, Sam L, Ronald, Charles, Marco S, Marco W, Sam Y, Enoch, Walter  Pre-Quiz 6  Tashi  SQ  Thursday (18 Feb 2016)  SHW7-01, 7-A1  Deadline: yesterday! 22 October 2015 Ronald HUI

Book 5B Chapter 7 Equations of Circles

y x O  P r  C(h, k) (x, y) Let (x, y) be the coordinates of P. PC = r By the distance formula Square both sides. The coordinates of all the points on the circle centred at C(h, k) with radius r satisfy this equation. rkh r k h

Therefore, the equation of the circle centred at C(h, k) with radius r is given by: This is known as the standard form or centre-radius form of the equation of a circle. h k r For example, the equation of the circle centred at C(0, 2) with radius 3 is given by (x – 0) 2 + (y – 2) 2 = 3 2 i.e. x 2 + (y – 2) 2 = 9

x 2 + y 2 + Dx + Ey + F = 0, where D, E and F are constants. General Form If we expand the left-hand side of the equation (x – h) 2 + (y – k) 2 = r 2, we have Let D = – 2h, E = – 2k and F = h 2 + k 2 – r 2. (x 2 – 2hx + h 2 ) + (y 2 – 2ky + k 2 ) = r 2 x 2 + y 2 – 2hx – 2ky + h 2 + k 2 – r 2 = 0 – 2k– 2h h 2 + k 2 – r 2 Then, In fact, the equation of any circle can be simplified to this form. We call this the general form of the equation of a circle. Note that, and.

(2)The coefficients of both x 2 and y 2 are equal to 1. x 2 + y 2 + Dx + Ey + F = 0 (1)The equation is quadratic, i.e. the highest degree of all its terms is 2. (3)The equation has no xy term. Do you notice any characteristics from the general form of the equation of a circle?

Follow-up question Determine whether each of the following equations represents a circle. If yes, put a '  ' in the box; otherwise, put a '  '. (a)x 2 + y 2 + 4y – 9 = 0 (b)3x 2 + 2y 2 + 3x + 2y + 8 = 0 (c)x 4 + y 4 + 4x + 7y – 7 = 0 (d)x 2 + y 2 – 2xy + 6y – 11 = 0    

Finding the Centre and Radius of a Circle Consider a circle with centre (h, k) and radius r. Standard form: (x – h) 2 + (y – k) 2 = r 2 General form: x 2 + y 2 + Dx + Ey + F = 0 D = –2h E = –2k F = h 2 + k 2 – r 2

Given the general form of the equation of a circle x 2 + y 2 + Dx + Ey + F = 0. Then

Example: For the circle x 2 + y 2 – 4x + 8y + 4 = 0, centre = |       2, 2 ED = 2 8, 2 )4(          )4,2(  = DEF radius

Follow-up question For each of the following equations of circles, find the centre and the radius of the circle. (a)x 2 + y x – 2y + 28 = 0 (b)4x 2 + 4y x – 4y + 33 = 0 (a)

Follow-up question (b) For each of the following equations of circles, find the centre and the radius of the circle. (a)x 2 + y x – 2y + 28 = 0 (b)4x 2 + 4y x – 4y + 33 = 0 Remember to convert the coefficients of x 2 and y 2 to 1.

If Consider the expression for the radius of the circle, i.e. r > 0 r > 0 r r This circle is a real circle. = 0 r = 0 This circle is a point circle. radius = 0 < 0 r is NOT a real number This circle is an imaginary circle. The circle does NOT exist. Nature of a Circle

It is given that A(–1, 4) and B(5, 6) are the end points of a diameter of a circle. Find the equation of the circle. Let C be the centre of the circle. ∵ C is the mid-point of AB. ∴ Coordinates of C

It is given that A(–1, 4) and B(5, 6) are the end points of a diameter of a circle. Find the equation of the circle. Let C be the centre of the circle. ∵ C is the mid-point of AB. ∴ Coordinates of C Radius of the circle ∴ The equation of the circle is (x – 2) 2 + (y – 5) 2 = 10.

It is given that O(0, 0), A(2, 2) and B(–4, –2) are points on a circle. Find the equation of the circle. Let x 2 + y 2 + Dx + Ey + F = 0 be the equation of the circle. ∵ O(0, 0), A(2, 2) and B(–4, –2) are points on a circle. ∴ By substitution, we have D(0) + E(0) + F = 0 F = (1) D(2) + E(2) + F = 0 2D + 2E + F = – (2) (–4) 2 + (–2) 2 + D(–4) + E(–2) + F = 0 4D + 2E – F = (3)

By substituting (1) into (2), we have 2D + 2E = –8 D + E = – (4) By substituting (1) into (3), we have 4D + 2E = 20 2D + E = (5) (5) – (4):D = 14 By substituting D = 14 into (4), we have 14 + E = –4 E = –18 ∴ The equation of the circle is x 2 + y x – 18y = 0.

Book 5B Chapter 7 More about Equations of Circles

Besides considering given points on a circle, the equation of the circle can also be found based on other given information. In the following, let me illustrate this with examples.

A circle with its centre lying on an axis In the figure, the circle passes through two points A(–3, –3) and B(–1, 1). The centre C of the circle lies on the y-axis. (a)Find the equation of the circle. (b)Determine whether the point P(4, –1) lies on, inside or outside the circle. In this case, we know two points on the circle and the centre of the circle lies on the y-axis. We can make use of the given conditions to find the centre and the radius of the circle.

A circle with its centre lying on an axis In the figure, the circle passes through two points A(–3, –3) and B(–1, 1). The centre C of the circle lies on the y-axis. (a)Find the equation of the circle. (b)Determine whether the point P(4, –1) lies on, inside or outside the circle. Since C lies on the y-axis, its x-coordinate must be 0. We can let (0, c) be the coordinates of C.

A circle with its centre lying on an axis In the figure, the circle passes through two points A(–3, –3) and B(–1, 1). The centre C of the circle lies on the y-axis. (a)Find the equation of the circle. (b)Determine whether the point P(4, –1) lies on, inside or outside the circle. CA and CB are two radii of the circle. Therefore, CA = CB. This can be used to find the y-coordinate of C.

(a) Let (0, c) be the coordinates of the centre C. ∵ CA = CB (radii) ∴ ∴ The coordinates of C are (0, –2). After finding the coordinates of C, we can find the radius by the distance formula. The radius of the circle ∴ The equation of the circle is x 2 + (y + 2) 2 = 10.

A circle with its centre lying on an axis In the figure, the circle passes through two points A(–3, –3) and B(–1, 1). The centre C of the circle lies on the y-axis. (a)Find the equation of the circle. (b)Determine whether the point P(4, –1) lies on, inside or outside the circle. How can we determine the position of P relative to the circle in part (b)?

A circle with its centre lying on an axis In the figure, the circle passes through two points A(–3, –3) and B(–1, 1). The centre C of the circle lies on the y-axis. (a)Find the equation of the circle. (b)Determine whether the point P(4, –1) lies on, inside or outside the circle. We can compare the distance between C and P with the radius r.

Case 1: P lies outside the circle. CP is longer than the radius of the circle. i.e. CP > r

Case 2: P lies on the circle. CP is equal to the radius of the circle. i.e. CP = r

Case 3: P lies inside the circle. CP is shorter than the radius of the circle. i.e. CP < r

(b) From (a), coordinates of C = (0, –2) and radius i.e. CP > radius ∴ P(4, –1) lies outside the circle. P(4, –1)

A circle touching two horizontal lines and one vertical line In the figure, the straight lines x = 3, y = 10 and y = –2 touch the circle at P, Q and R respectively. (a)Find the radius of the circle. (b)Hence, find the equation of the circle. Without the centre, I may find a diameter first. Is QR a diameter of the circle?

Let C be the centre. C tangent ⊥ radius ∴ QCR is a vertical line segment. ∴ QR is a diameter of the circle. ∵ y = 10 and y = –2 are two horizontal lines.

C (a)Radius of the circle

C (b)Let (a, b) be the coordinates of C. (a, b) 3 radius a = 3 + radius = = 9 = 6

C (9, b) 10 radius b = 10 – radius = 10 – 6 = 4 = 6

C (9, 4) radius = 6 ∴ The centre is C(9, 4). ∴ The equation of the circle is (x – 9) 2 + (y – 4) 2 = 36

A circle with its centre lying on a given straight line In the figure, the circle cuts the x-axis at two points P and Q. Its centre C lies on the straight line L: 2x + y – 15 = 0. It is given that the radius of the circle is and PQ = 4. Find (a)the centre of the circle, (b)the equation of the circle. How can I use the given information to find the coordinates of C?

M M is the mid-point of PQ, PQ ⊥ CM. The length of CM is the y-coordinate of C. Let (h, k) be the coordinates of C and M be the mid-point of PQ.

M △ CMP is a right- angled triangle. The lengths of CP and PM are known. The length of CM can be found by Pyth. Theorem. Let (h, k) be the coordinates of C and M be the mid-point of PQ.

M ∵ PM = QM ∴ CM ⊥ PQ (line joining centre to mid-pt. of chord ⊥ chord) Consider △ CMP. i.e. k = 1

Note that the centre C lies on the straight line L. If the x-coordinate or the y-coordinate of C is known, the other coordinate of C can be found.

∵ C(h, 1) lies on the straight line L: 2x + y – 15 = 0. ∴ 2h + 1 – 15 = 0 h = 7 (b)The equation of the circle is ∴ The centre of the circle is (7, 1).