# The Circle (x 1, y 1 ) (x 2, y 2 ) If we rotate this line we will get a circle whose radius is the length of the line.

## Presentation on theme: "The Circle (x 1, y 1 ) (x 2, y 2 ) If we rotate this line we will get a circle whose radius is the length of the line."— Presentation transcript:

The Circle (x 1, y 1 ) (x 2, y 2 )

If we rotate this line we will get a circle whose radius is the length of the line

Hence the equation of a circle, centre the origin is

b) Show that the point A(3,4) lies on the circle x 2 + y 2 = 25 When x = 3 and y = 4, x 2 + y 2 = 3 2 + 4 2 = 25 Since the point (3,4) satisfies the equation it must lie on the circle.

c) check that the point B(3,3) lies inside the circle and that C(3,5) lies outside the circle with equation x 2 + y 2 = 25 When x = 3 and y = 3,x 2 + y 2 = 3 2 + 3 2 = 18 As 18 < 25, (3,3) must lie inside the circle. When x = 3 and y = 5,x 2 + y 2 = 3 2 + 5 2 = 34 As 34 > 25, (3,5) must lie outside the circle.

Circles with other Centres (a, b) Here the centre has moved from (0,0) to (a,b). The equation of a circle centre (a,b) Radius r is (x-a) 2 + (y-b) 2 = r 2

(x - a) 2 + (y - b) 2 = r 2 (x - 1) 2 + (y - (-2)) 2 = (4  3) 2 (x - 1) 2 + (y + 2) 2 = (4  3) × (4  3) (x - 1) 2 + (y + 2) 2 = 48 Or in expanded form... (x - 1)(x - 1) + (y + 2)(y + 2) = 48 x 2 - 2x + 1 + y 2 + 4y + 4 = 48 x 2 + y 2 - 2x + 4y - 43 = 0

b)P(2,-2) and Q(8,6) are points on a circle such that PQ is a diameter. Find i) The coordinates of the centre ii) Its radius iii) Its equation (i) Centre is mid point of diameter Mid point of PQ is (ii) Distance from (5,2) to (8,6) = 5 units

(iii) Centre (5,2) radius 5 units

The General Equation of a Circle This is the general equation of a circle centre (-g, -f) and radius

Example Find the radius and centre of the circle with equation x 2 + y 2 + 6x - 8y + 24 = 0 Equating with x 2 + y 2 + 2gx + 2fy + c = 0 It follows that g = 3, f = -4 and c = 24 Centre = (-g,-f)

Intersection of a Line and a Circle A circle and a line can have 2 points of intersection 1 point of intersection No points of intersection

Example Find the point(s) of intersection of the line 5y - x + 7 = 0 and the circle with equation x 2 + y 2 + 2x - 2y - 11 = 0 i) Rearrange 5y - x + 7 = 0 into form x =… or y =... So that we can substitute it into the equation of the circle. x = 5y + 7. Now substitute this into the circle equation. ( 5y + 7 ) 2 + y 2 + 2( 5y + 7 ) - 2y - 11 = 0 25y 2 +70y+ 49 + y 2 + 10y + 14 - 2y - 11 = 0 26y 2 +78y + 52 = 0 26(y 2 +3y + 2) = 0 (y 2 +3y + 2) = 0 (y + 1)(y + 2) = 0 y = -1 or y = -2

Since x = 5y + 7, when y = -1 x = 2 when y = -2 x = -3 Points of contact (2, -1) and (-3, -2)

Tangents to Circles When a line meets a circle at one point only, then the line is a tangent to the circle.

a) Prove that the line x - 7y - 50 = 0 is a tangent to the circle with equation x 2 + y 2 = 50 There are two methods we can use to answer this question. (i)Solve the equation as before and show 1 point of contact (ii)Use the discriminant to show tangency using the resulting equation after substitution.

a) Prove that the line x - 7y - 50 = 0 is a tangent to the circle with equation x 2 + y 2 = 50 Method 1. Substituting gives: Since only one value for y is produced then x - 7y - 50 = 0 is a tangent b 2 - 4ac = 14 2 - (4 × 1 × 49) Method 2. = 0 Since b 2 - 4ac = 0 then there is only one point of contact so x - 7y - 50 = 0 is a tangent to the circle.

Equations of Tangents The equation of a tangent at a point on a circle can be found from the point of contact Using m tan × m radius = -1

a) Find the equation of the tangent from the point (0,3) to the circle with equation x 2 + y 2 - 8 x -2 y - 3 = 0 The equation of the tangent must be Substituting into the circle equation gives: x 2 +(mx + 3) 2 - 8 x -2(mx + 3) - 3 = 0 x 2 + m 2 x 2 + 6mx + 9 - 8 x - 2mx - 6 - 3 = 0 x 2 + m 2 x 2 + 4mx - 8 x = 0 x 2 (1+ m 2 ) + x (4m - 8 ) = 0 For tangency b 2 - 4ac = 0 (4m - 8 ) 2 - 4 (1+ m 2 ) × 0 = 0 16m 2 - 64m + 64 = 0 16(m 2 - 4m + 4) = 0

16(m - 2)(m - 2) = 0 (m - 2)(m - 2) = 0 m = 2 The equation of the tangent is y = mx + 3

b) Show that A (8,3) lies on the circle x 2 + y 2 - 8 x -2 y - 3 = 0 and hence find the equation of the tangent at A. Centre of the circle =(4,1) Equation of the tangent = When x = 8 and y = 3,x 2 + y 2 - 8 x -2 y - 3 = 8 2 + 3 2 –64-6-3 = 0 Hence (8,3) lies on the circle.

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