23/09/2017 EQUATION OF A CIRCLE ANALYTICAL GEOMETRY ON CENTRE The general equation of a circle on centre (0;0) P(x;y) i.e. the ORIGIN is r x2 + y2 = r 2 r where P (x; y) is a point on the circle Recap: Terminology of a Circle and r is the radius EXAMPLE … EQUATION OF A CIRCLE OFF CENTRE Find the equation of a circle that passes Let’s investigate the equation of a circle through the origin and the point (-1; 4). off centre, using Pythagoras: P(x;y) 2 2 2 x + y = r PD = y – y1 2 2 2 r ( -1) + (4) = r CD = x – x1 C (x ;y ) 2 1 1 r = 17 2 2 2 2 2 r = PD + CD D(x; y1) equation is x + y = 17 2 2 2 r = (y – y1) + (x – x1) NB the radius is 17 EQUATION OF A CIRCLE EQUATIONS OF CIRCLES OFF CENTRE The general equation of a circle Investigating Circles OFF centre (i.e. NOT on the P(x;y) origin) is r Equations of circles on- and off-origin (x – a)2 + (y – b)2 = r 2 Determining the centre and radius of a circle Completing the square to write circles in standard where O (a; b) form P (x; y) is a point on the circle O (a; b) is the centre of the circle Practise completing the square to determine the centre and radius of a circle and r is the radius 1 23/09/2017 EXAMPLE … EXAMPLE … Find the general equation a circle centre Determine the equation the circle, given P(x;y) (-1;3) with a radius of 5 units. the diameter AB: Midpoint! A(2;6) r = 5 (x – a)2 + (y – b)2 = r 2 Determine the centre… C(-1;3) 푥 + 푥 푦 + 푦 ( x – (-1) )2 + (y – 3) 2 = (5)2 1 2 ; 1 2 2 2 x 2 +2x+2 + y2 -6y +9 = 25 2 − 4 6 − 2 ; 2 2 x 2 +2x + y2 - 6y = 14 ( −1; 2) B(-4;-2) EXAMPLE continued… EXAMPLE continued… Determine the radius… Substitute a Centre: −1; 2 point! A(2;6) Subst. A(2;6) A(2;6) Therefore … (x + 1 )2 + (y – 2)2 = r 2 (x – a)2 + (y – b)2 = r 2 (2 + 1 )2 + (6 – 2)2 = r 2 ( x – (-1) )2 + (y – 2) 2 = r 2 2 2 2 2 r = 25 (x + 1 ) + (y – 2) = r B(-4;-2) ∴ (x + 1 )2 + (y – 2)2 = 25 B(-4;-2) PRACTISE! TANGENTS TO CIRCLES A straight line can intersect a circle, i.e. Sketching Circles with Origin Off-Centre Find the Equations of Circles Visually ONCE TWICE NOT cut A tangent A secant the circle Identifying Tangents 2 23/09/2017 TANGENTS TO CIRCLES EXAMPLE … The relationship between perpendicular Show that 2x + 5 + y = 0 distance from origin and the length of the radius Solve is a tangent to the circle simultaneously! M M O x2 + y2 = 5 O O y = - 2x – 5 (i) x2 + y2 = 5 (ii) M Substitute (i) into (ii) Cut TWICE NOT cut Cut ONCE 2 2 A tangent A secant the circle x + (- 2x – 5) = 5 OM > r OM = r OM < r EXAMPLE continued… TANGENTS TO CIRCLES x2 + (- 2x – 5)2 = 5 • If 2 DIFFERENT points of x2 + (4x2 +20x + 25) = 5 intersection are found, the line is a Since there is SECANT. 2 5x +20x + 20 = 0 only 1 point of x2 + 4x + 4 = 0 intersection • If there is no solution, the line does (i.e. x= -2), the not intersect the circle. (x + 2)(x + 2) = 0 line is a x = -2 or x = -2 TANGENT Intersection of Lines and Circles EXAMPLE … EXAMPLE continued… Find the equation of the tangent to the ∴ 푦 = 2푥 + 푐 Now solve for c! circle x2 + y2 = 5 at the point ( – 2; 1). Substitute point of contact of tangent: Tangent: y = mx + c Solve for m first! 1 − 0 −1 Subst. P(-2; 1) 푚푟푎푑푖푢푠 = = P(-2;1) 1 = 2 −2 + 푐 P(-2;1) −2 − 0 2 r r O(0; 0) 푐 = 5 O(0; 0) Radius is to Tangent 푚푟푎푑푖푢푠 × 푚푡푎푛푔푒푛푡 = −1 ∴ 퐸푞푢푎푡푖표푛 표푓 푡푎푛푔푒푛푡 푖푠: ∴ 푚 = 2 푡푎푛푔푒푛푡 푦 = 2푥 + 푐 Working with tangents to circles 3 23/09/2017 EXAMPLE … EXAMPLE continued… −1 Find the equation of the tangent at the ∴ 푦 = 푥 + 푐 point P(1;3), given the centre C(-1;-1). 2 P(1;3) r P(1;3) Tangent: y = mx + c r Subst. P(1; 3) 3 − (−1) −1 C(-1;-1) 푚 = = 2 3 = 1 + 푐 푟푎푑푖푢푠 1 − (−1) C(-1;-1) 2 7 푐 = 푚푟푎푑푖푢푠 × 푚푡푎푛푔푒푛푡 = −1 2 −1 −1 7 ∴ 푚 = (radius tangent) ∴ 퐸푞푢푎푡푖표푛 표푓 푡푎푛푔푒푛푡: 푦 = 푥 + 푡푎푛푔푒푛푡 2 2 2 EXAMPLE … EXAMPLE continued… 4 Find the equation of the tangent to the ∴ 푦 = 푥 + 푐 circle (x – 2)2 + (y - 3)2 = 16 at (6;0). 3 Find the Equation Subst. P(6; 0) of a Tangent to a Centre: (2;3) What is the 4 Circle 3 − 0 −3 centre of the circle? 0 = 6 + 푐 푚푟푎푑푖푢푠 = = 3 GeoGebra: 2 − 6 4 Analytical 푐 = −8 Geometry Tool 푚 × 푚 = −1 (radius tangent) 푟푎푑푖푢푠 푡푎푛푔푒푛푡 ∴ 퐸푞푢푎푡푖표푛 표푓 푡푎푛푔푒푛푡 푖푠: 4 Complex Example 4 of Tangents to ∴ 푚푡푎푛푔푒푛푡 = 푦 = 푥 + 8 3 3 Circle EXAMPLE … EXAMPLE continued… Given the circle with diameter AB and b) The equation of the circle centre (1;-1), determine: 2 2 2 Midpoint in reverse! (x – a) + (y – b) = r a) The c0-ordinates of B ( x – 2 )2 + (y – (-3)) 2 = r 2 푥+5 푦+0 = 2 = −3 Subst. A(5; 0) 2 2 푥 + 5 = 4 푦 + 0 = −6 (5 - 2 )2 + (0 + 3)2 = r 2 푥 = −1 푦 = −6 r 2 = 18 ∴ B( −1; −6) ∴ (x - 2 )2 + (y + 3)2 = 18 4 23/09/2017 EXAMPLE continued… EXAMPLE continued… c) The equation of the tangent at A c) Determine the equation of a line parallel 0 − (−3) to the tangent at A and passing through the 푚푟푎푑푖푢푠 = = 1 5 − 2 point C (2;8). ∴ 푚 = −1 푡푎푛푔푒푛푡 Tangent at A: 푦 = −푥 + 5 (radius tangent) Substitute point A Parallel line: 푦 = −푥 + 푐 푦 = −푥 + 푐 8 = −(2) + 푐 0 = − 5 + 푐 푐 = 10 Substitute 푐 = 5 point C ∴ 푦 = −푥 + 5 ∴ 푦 = −푥 + 10 EXAMPLE continued… EXAMPLE continued… d) Determine whether the tangent at B will Tangent at B −1; −6 : Tangent at A: intersect with the tangent at A. 푦 = −푥 + 푐 푦 = −푥 + 5 Tangent at A: 푦 = −푥 + 5 −6 = − −1 + 푐 푐 = −7 Tangent at B −1; −6 : ∴ 푦 = −푥 − 7 −6 − (−3) 푚 = = 1 푟푎푑푖푢푠 −1 − 2 ∴ −푥 + 5 = −푥 − 7 ∴ 푚 = −1 푡푎푛푔푒푛푡 0 = −12 푁표 푠표푙푢푡푖표푛 … (radius tangent) 푇푎푛푔푒푛푡푠 푑표 푛표푡 푖푛푡푒푟푠푒푐푡 5 .