C2 Chapter 4: Equations of Circles Dr J Frost Last modified: 10 th September 2015

Midpoint of two points and distance between them (1,1) (2,2)(1.5, 1.5) √2√2 MidpointDistance (1,2) (4,6) 5 (2.5, 4) (a,a-1) (b,a+1)([a+b]/2, a) √ [(b-a) 2 +4] (1,-4) (-3,5)(-1, 0.5) √ 97 (-1,-2) (-2,10)(-1.5, 4) √ 145 ?? ?? ?? ?? ??

Centres of circles ? ?

Radii of circles The line PQ is a diameter of a circle, where P and Q are (-1, 3) and (6, -3) respectively. Find the radius of the circle. ? The points (-3, 19), (-15, 1) and (9, 1) are points on the circumference of a circle. Show that (-3, 6) is the centre of the circle. We just need to show that each point is equidistant from the centre. The distance for all three points to the centre turns out to be 13. ?

Equation of a circle x y x 2 + y 2 = r 2 ?

Equation of a circle x y Now suppose we shift the circle so it’s now centred at (a,b). What’s the equation now? (Hint: What would the sides of this right- angled triangle be now?) (x-a) 2 + (y-b) 2 = r 2 ? r

Equation of a circle CentreRadiusEquation (0,0)5x 2 + y 2 = 25 (1,2)6(x-1) 2 + (y-2) 2 = 36 (-3,5)1(x+3) 2 + (y-5) 2 = 1 (-5,2)7(x+5) 2 + (y-2) 2 = 49 (-6,-7)4(x+6) 2 + (y+7) 2 = 16 (1,-1) √3√3 (x-1) 2 + (y+1) 2 = 3 ? ? ?? ?? ?? ?? (-2,3) 2√ 2 (x+2) 2 + (y-3) 2 = 8 ??

Test Your Understanding C2 Jan 2005 Q2 ? ?

Completing the square [Edexcel] The circle C, with centre at the point A, has equation x 2 + y 2 – 10x + 9 = 0. Find (a)the coordinates of A(2) (b)the radius of C,(2) x 2 – 10x + y = 0 (x – 5) 2 – 25 + y = 0 (x – 5) 2 + y 2 = 16 So centre is (5,0), radius is 4. ?

Equation of a circle EquationCentreRadius x 2 + y 2 + 4y – 5 = 0(0, -2)3 ? x 2 + y 2 – 6x + 4y – 3 = 0(3, -2)4 x 2 + y x + 2y + 12 = 0(-6, -1)5 ? ? ? ? ? (Note: this appears in exams, but not in your textbook!) x 2 + y 2 – 4x – 6y = 3(2, 3)4 ?? x 2 + y 2 + x + y = 1(-0.5, -0.5) ??

Lines through circles There are two types of questions you can get in C2 exams where you’d have to determine the equation of a line, based on two different circle theorems: The perpendicular from the centre of a circle to a chord bisects the chord. We say that the radius is the perpendicular bisector of the chord. The perpendicular from the centre of a circle to a chord bisects the chord. How do you think we might find the equation of the perpendicular bisector? Find midpoint of chord and use negative reciprocal of gradient. How do you think we might find the equation of the tangent at a given point? Use negative reciprocal of the gradient of the radius combined with the given point of contact. ? ? ? ?

Examples The line AB is a diameter of the circle centre C, where A and B are (-1, 4) and (5,2) respectively. The line l passes through C and is perpendicular to AB. Find the equation of l. y = 3x - 3 ? The line PQ is a chord of the circle centre (-3, 5), where P and Q are (5, 4) and (1, 12) respectively. The line l is perpendicular to PQ and bisects it. Prove that l passes through the centre of the circle. ?

Test Your Understanding ? D C B A x y ?

More Relevant Circle Theorems A B C ? Tangents from a point to a circle are equal in length. ? ?

Test Your Understanding ?

Questions Ex4A Q10) The points V(-4, 2a) and W(3b, -4) lie on the circle centre (b, 2a). The line VW is a diameter of the circle. Find the value of a and b. a = -2, b = 4 Ex4B Q1) The line FG is a diameter of the circle centre C, where F and G are (-2, 5) and (2, 9) respectively. The line l passes through C and is perpendicular to FG. Find the equation of l. y= -x + 7 Ex4B Q9) The points P(3, 16), Q(11, 12) and R(-7, 6) lie on the circumference of a circle. a)Find the equation of the perpendicular bisector of: i.PQ ii.PR b)Hence, find the coordinates of the centre of the circle. a) i) y = 2x ii) y = -x + 9 b) (3, 6) Ex4B Q10) Find the centre of a circle with points on its circumference of (-3,19), (9,11) and (-15, 1). (-3, 6) Ex4C Q9) The points A(2, 6), B(5,7) and C(8,- 2) lie on a circle. a)Show that triangle ABC has a right angle. b)Find the area of the triangle. c)Find the centre of the circle. a)AB = √ 10, AC = 10. BC = r90. AB 2 + BC 2 = AC 2. b)15 c)(5, 2) ? ? ? ? ? ? ? (For Further Mathematicians give exam questions at this point)

Equation of a circle Show that the circle (x-3) 2 + (y+4) 2 = 20 passes through (5, -8). Just substitute x=5 and y=-8 and show the equation holds! Given that AB is the diameter of a circle where A and B are (4,7) and (-8,3) respectively, find the equation of the circle. (x+2) 2 + (y-5) 2 = 40 The line 4x – 3y – 40 = 0 is a tangent to the circle (x – 2) 2 + (y – 6) 2 = 100 at the point P(10,0). Show that the radius at P is perpendicular to the line (as we would expect!) P Equation of line is y = 4/3 x – 40/3, so gradient is 4/3 Centre of circle is (2,6) So gradient of radius is (6-0)/(2-10) = -3/4. Since 4/3 x -3/4 = -1, radius is perpendicular to tangent. ? ? ? Where does the circle (x – 1) 2 + (y – 3) 2 = 45 meets the x-axis? On x-axis, y = 0. So (x – 1) = 45. Solving gives (7,0) and (-5,0) ? ? ? ? ?

Intersections! How could you tell if a line and a circle intersect: 0 times twice once Equate the expressions then look at the discriminant: b 2 – 4ac < 0 b 2 – 4ac > 0b 2 – 4ac = 0 ? ?? (x+2) 2 + y 2 = 33 y = x - 7 This allows us to prove that a line is a tangent to the circle.