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Mat 104 Elementary Mathematics Iii 3 Credits Co MAT 104 ELEMENTARY MATHEMATICS III 3 CREDITS CO-ORDINATE GEOMETRY (CONT’D) CIRCLES The equation of a circle with centre (푎, 푏) and radius 푟 is given by (푥 − 푎)2 + (푦 − 푏)2 = 푟2 When the circle has its centre at origin, the equation of the circle becomes or reduces to 푥2 + 푦2 = 푟2 Because, circle with centre at origin means the coordinates of the centre is (0, 0) ⟹ (푥 − 푎)2 + (푦 − 푏)2 = 푟2 (푥 − 0)2 + (푦 − 0)2 = 푟2 푥2 + 푦2 = 푟2 Examples 1. Find the equation of the circle which has its center at origin and a radius of 5. Solution Simply from 푥2 + 푦2 = 푟2 ⟹ 푥2 + 푦2 = 52 Thus 푥2 + 푦2 = 25 2. Find the equation of the circle with center (2, 5) and a radius of 3. Solution From (푥 − 푎)2 + (푦 − 푏)2 = 푟2 ⟹ (푥 − 2)2 + (푦 − 5)2 = 32 ⟹ 푥2 + 푦2 − 4푥 − 10푦 + 20 = 0 3. Find the equation of a circle with centre (−2, 3) which passes through the point (1, 2) Solution Length of radius, 푟 = √(−2 − 1)2 + (3 − 2)2 푟 = √−32 + 12 푟 = √10 푟2 = 10 Using (푥 − 푎)2 + (푦 − 푏)2 with the circle centre (−2, 3) and radius 푟 = √10 The equation of the circle is (푥 + 2)2 + (푦 − 3)2 = 10 ⟹ 푥2 + 푦2 + 4푥 − 6푦 + 3 = 0 4. Find the radius and centre of the circle whose equation is given by 3푥2 + 3푦2 − 24푥 + 12푦 + 11 = 0 Solution 11 푥2 + 푦2 − 8푥 + 4푦 + = 0 3 2 2 11 푥 + 푦 − 8푥 + 4푦 = − 3 Completing squares 11 푥2 − 8푥 + 16 + 푦2 + 4푦 + 4 = − + 16 + 4 3 2 2 49 (푥 − 4) + (푦 + 2) = 3 49 Comparing with (푥 − 푎)2 + (푦 − 푏)2 = 푟2 ⟹ 푎 = 4, 푏 = −2, 푟 = √ 3 7 Circle centre (4, −2) and radius √3 EQUATION OF A CIRCLE GIVEN INFORMATION ABOUT THE DIAMETER The equation of a circle can also be deduced if the information about the points which the diameter passes through is known. In summary the equation of a circle whose diameter is the join of (푥1, 푥2) and (푦1, 푦2) is given by the equation (푥 − 푥1)(푥 − 푥2) + (푦 − 푦1)(푦 − 푦2) = 0 Examples 1. Find the equation of a circle drawn on diameter 퐴퐵 where 퐴 is (−1, 3) and 퐵 is (3, 2) Solution 5 Mid-point of 퐴퐵; (1, ) ⟹ centre of circle 2 5 Thus, 푟2 = (1 + 1)2 + ( − 3)2 2 1 푟2 = 22 + (− )2 2 17 푟2 = 4 (Note that to get the length of 푟, you can use any of the two points with the 5 5 circle centre i.e. {(−1, 3) and (1, )} or {(3, 2) and (1, )} 2 2 2 5 2 17 Equation of circle (푥 − 1) + (푦 − ) = 2 4 25 17 푥2 − 2푥 + 1 + 푦2 − 5푦 + = 4 4 푥2 + 푦2 − 2푥 − 5푦 + 3 = 0 Using the formula method (푥 − 푥1)(푥 − 푥2) + (푦 − 푦1)(푦 − 푦2) = 0 ⟹ (푥 + 1)(푥 − 3) + (푦 − 3)(푦 − 2) = 0 ⟹ (푥2 − 2푥 − 3) + (푦2 − 5푦 + 6) = 0 ⟹ 푥2 + 푦2 − 2푥 − 5푦 + 3 = 0 2. The equation of a circle is 푥2 + 푦2 − 6푥 + 8푦 = 24. If the circle which passes through the point (−1, 1), find the coordinates of the other end of the diameter and the equation of the diameter. Solution 푥2 + 푦2 − 6푥 + 8푦 = 24 푥2 − 6푥 + 9 + 푦2 + 8푦 + 16 = 24 + 9 + 16 (푥 − 3)2 + (푦 + 4)2 = 49 Thus the circle centre is (3, −4) ⟹ 푚푖푑푝표푖푛푡 표푓 푡ℎ푒 푑푖푎푚푒푡푒푟 Note that the first end of the diameter is (−1, 1), so the coordinates of the other end of the diameter is obtained as 1 1 3 = (푥 + 푥 ) −4 = (푦 + 푦 ) 2 1 2 2 1 2 1 1 3 = (−1 + 푥 ) −4 = (1 + 푦 ) 2 2 2 2 푥2 = 7 푦2 = −9 Thus the coordinates of the other end of the diameter is (7, −9). Now since the coordinates of the two points of the diameter is known (−1, 1) and (7, −9). the equation of the line (i.e. the diameter) is obtained using the two-point form 푦2− 푦1 푦 − 푦1 = (푥 − 푥1) 푥2− 푥1 −9 −1 푦 − 1 = (푥 + 1) 7+ 1 5 푦 − 1 = − (푥 + 1) ⟹ 4푦 + 5푥 + 9 = 0 4 Alternatively, To find the equation of the diameter, we can use just the coordinates of the point that we were given i.e. (−1, 1) and the coordinates of the circle centre i.e. (3, −4), which is equally the mid-point of the diameter. Using the two-point form, 푦2− 푦1 푦 − 푦1 = (푥 − 푥1) 푥2− 푥1 −4 −1 푦 − 1 = (푥 + 1) 3 + 1 5 푦 − 1 = − (푥 + 1) ⟹ 4푦 + 5푥 + 9 = 0 4 (The reason for this is because the three points – the two ends of the diameter and the circle centre i.e. (−1, 1), (7, −9) and (3, −4) lie on the same straight line and as such are collinear points meaning they will always have the same gradient). GENERAL FORM OF THE EQUATION OF A CIRCLE The equation of a circle with centre (푎, 푏) and radius 푟 is (푥 − 푎)2 + (푦 − 푏)2 = 푟2 which gives 푥2 + 푦2 − 2푎푥 − 2푏푦 + 푎2 + 푏2 − 푟2 = 0 written in the form 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0, which is the general standard form of the equation of a circle. Any of 푔, 푓 and 푐 can be 0 With This general form of the equation of a circle, the equation of a circle can be determined given that it passes through three points by substituting the (푥, 푦) values of each point into the equation above and solving the three equations simultaneously. Examples 1. Find the equation of a circle through the points (6, 1), (3,2) and (2, 3) Using 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0 For the point (6, 1), we have 62 + 12 + 2(6)푔 + 2(1)푓 + 푐 = 0 (푖) For the point (3, 2), we have 32 + 22 + 2(3)푔 + 2(2)푓 + 푐 = 0 (푖푖) For the point (2, 3), we have 22 + 32 + 2(2)푔 + 2(3)푓 + 푐 = 0 (푖푖푖) This becomes 37 + 12푔 + 2푓 + 푐 = 0 (푖푣) 13 + 6푔 + 4푓 + 푐 = 0 (푣) 13 + 4푔 + 6푓 + 푐 = 0 (푣푖) Subtract equation (푣푖) from (푣), we have 2푔 − 2푓 = 0 ⟹ 푔 = 푓 (푣푖푖) Subtract equation (푣) from (푖푣), we have 24 + 6푔 − 2푓 = 0 (푣푖푖푖) Substitute for 푔 in equation (푣푖푖푖) 24 + 6푓 − 2푓 = 0 (since 푔 = 푓) 4푓 = −24 푓 = −6 ⟹ 푔 = −6 Solving for 푐, we have 푐 = 47 Thus the equation of the circle is 푥2 + 푦2 − 12푥 − 12푦 + 47 = 0 2. Find the equation of a circle through the points (0, 0), (3, 1) and (5, 5) Using 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0 For the point (0, 0), we have 02 + 02 + 2(0)푔 + 2(0)푓 + 푐 = 0 ⟹ 푐 = 0 (푖) For the point (3, 1), we have 32 + 12 + 2(3)푔 + 2(1)푓 + 푐 = 0 ⟹ 10 + 6푔 + 2푓 + 푐 = 0 (푖푖) For the point (5, 5), we have 52 + 52 + 2(5)푔 + 2(5)푓 + 푐 = 0 ⟹ 50 + 10푔 + 10푓 + 푐 = 0 (푖푖푖) Solve equations (푖푖) and (푖푖푖) simultaneously, we have 푔 = 0 and 푓 = −5 Thus the equation of the circle is 푥2 + 푦2 − 10푦 = 0 CONCYCLIC POINTS Four or more points are said to be concyclic if they lie on the same circle. Example Determine if the points (5, 2), (2, 3), (−3, −2) and (6, −5) are concyclic. Solution Using 푥2 + 푦2 + 2푔푥 + 2푓푦 + 푐 = 0 For the point 52 + 22 + 2(5)푔 + 2(2)푓 + 푐 = 0 (푖) For the point 22 + 32 + 2(2)푔 + 2(3)푓 + 푐 = 0 (푖푖) For the point −32 + −22 + 2(−3)푔 + 2(−2)푓 + 푐 = 0 (푖푖푖) Solving equations (푖), (푖푖) and (푖푖푖) simultaneously, 푔 = −2, 푓 = 2, 푐 = −17 Thus the equation of the circle is 푥2 + 푦2 − 4푥 + 4푦 − 17 = 0 Now if the points are concyclic input the coordinates of the fourth point into the equation of the circle to verify if this fourth point satisfies the equation Therefore, we put the point (6, −5) into the equation 푥2 + 푦2 − 4푥 + 4푦 − 17 = 0 We have, 62 + (−5)2 − 4(6) + 4(−5) − 17 = 0 ⟹ 36 + 25 − 24 − 20 − 17 = 0 This shows that the points are concyclic. POINTS INSIDE OR OUTSIDE A CIRCLE Consider the function 푓(푥, ) = (푥 − 푎)2 + (푦 − 푏)2 − 푟2. 푓(푥, 푦) = 0 defines a circle with radius 푟 and centre (푎, 푏). If the point (ℎ, 푘) lies inside the circle then (ℎ − 푎)2 + (푘 − 푏)2 < 푟2 i. e. (ℎ − 푎)2 + (푘 − 푏)2 − 푟2 < 0 or 푓(ℎ, 푘) < 0. So if (ℎ, 푘) lies inside the circle given by 푓(푥, 푦) = 0, then 푓(ℎ, 푘) < 0 and if the point (ℎ, 푘) lies outside the circle then 푓(ℎ, 푘) > 0. Example 1. Are the points 퐴 is (−1, −1) and 퐵 is (5, 2) inside or outside the circle 푥2 + 푦2 − 3푥 + 4푦 = 12 . Let 푓(푥, 푦) = 푥2 + 푦2 − 3푥 + 4푦 − 12 푓(1, −1) = 1 + 1 − 3 − 4 − 12 = −17 < 0 ⟹ 퐴 is inside the circle 푓(5, 2) = 25 + 4 − 15 + 8 − 12 = 10 > 0 ⟹ 퐵 is outside the circle POINT OF INTERSECTION OF A STRAIGHT LINE AND A CIRCLE The coordinates of the point of intersection of a straight line and a circle is obtained by solving the equation of the line and the circle simultaneously.
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