Block 1

Circles and Tangents

What is to be learned?

• How to find the equation of a tangent to a circle

Some revisionm1m2 = -1

For equation need m (a , b)

use y – b = m(x – a)

x2 + y2 + 2x – 4y – 5 = 0

Equation of Tangent at (2 , 1)

get centre

x2 + y2 + 2gx + 2fy + c = 0

2g = 2 , 2f = -4

g = 1 , f = -2

Centre (-1 , 2)

Equation of Tangent at (2 , 1)

Centre (-1 , 2)

(2 , 1)

(-1 , 2)

Need

(a , b)

m

m = 1 – 2 2 – (-1)

= -1/3

so m of tangent = 3

use y – b = m(x – a)

x2 + y2 – 6x – 14y – 3 = 0

Equation of Tangent at (5 , 8)

get centre

x2 + y2 + 2gx + 2fy + c = 0

2g = -6 , 2f = -14

g = -3 , f = -7

Centre (3 , 7)

Equation of Tangent at (5 , 8)

Centre (3 , 7)

(5 , 8)

(3 , 7)

Need

(a , b)

m

m = 7 – 8 3 – 5

= ½

so m of tangent = -2

y – b = m(x – a) m = -2, (a , b) = (5 , 8)y – 8 = -2(x – 5)y – 8 = -2x + 10

y = -2x + 18

Equation of a Tangent to a Circle

If given circle equation and point where tangent meets circle

Tactics• From equation find centre• Find gradient of radius• Use this to get gradient of tangent• Get equation of tangent

(m1m2 = -1)y – b = m(x – a)

x2 + y2 + 2x – 6y + 8 = 0

Equation of Tangent at (1 , 2)

get centre

x2 + y2 + 2gx + 2fy + c = 0

2g = 2 , 2f = -6

g = 1 , f = -3

Centre (-1 , 3)

Equation of Tangent at (1 , 2)

Centre (-1 , 3)

(1 , 2)

(-1 , 3)

Need

(a , b) m

(1 , 2)

m = 2 – 3 1 – (-1)

= -1/2

so m of tangent = 2

2

Now use y – b = m(x – a)

x2 + y2 + 12x – 10y – 3 = 0

Find equation of Tangent at (-2 , 4)

get centrex2 + y2 + 2gx + 2fy + c = 0

2g = 12 , 2f = -10

g = 6 , f = -5

Centre (-6 , 5)

Key Question

Equation of Tangent at (-2 , 4)

Centre (-6 , 5)

(-2 , 4)

(-6 , 5)

Need

(a , b)

m

m = 5 – 4 -6 + 2

= -¼

so m of tangent = 4

y – b = m(x – a) m = 4, (a , b) = (-2 , 4)y – 4 = 4(x + 2)y – 4 = 4x + 8

y = 4x + 12

Intersections of Lines and Circles

What is to be learned?

• How to find the point(s) of intersection between line and circleIf any!

Point of Intersection

x2 + y2 = 5

y = 2x + 5

simultaneous equations

y = y

substitution

Point of Intersection

x2 + y2 = 5

y = 2x + 5

Point of Intersection

x2 + y2 = 5

y = 2x + 5

x2 + (2x + 5)2 = 5

(2x + 5)2 =(2x + 5)(2x + 5) =4x2 + 20x + 25

= x2 + 4x2 + 20x +25 = 5= 5x2 + 20x +20 = 0= 5(x2 + 4x + 4) = 0= 5(x + 2)(x + 2) = 0

x = -2 y?y= 2(-2) + 5 = 1

(-2 , 1)

One Point of Intersection Tangent

No solution ifb2 – 4ac < 0

Intersection of Line and Circle

• Must use substitution• If only one point of intersection

Tangent• If no points of intersection

b2 – 4ac < 0

x2 + y2 + 2x – 2y – 11 = 0

5y = x – 7 x = 5y + 7 sub for x

x2 + y2 + 2x – 2y – 11 = 0

5y = x – 7 x = 5y + 7

(5y + 7)2 + y2 + 2(5y + 7) – 2y – 11 = 0 25y2 + 70y + 49 + y2 + 10y + 14 – 2y – 11 = 026y2 + 78y + 52 = 0

26(y2 + 3y + 2) = 0

26(y + 2)(y + 1) = 0

y = -2 or -1

y = -2 x = 5(-2) + 7

y = -1 x = 5(-1) + 7

(-3 , -2)

(2 , -1)

*

*(5y + 7)(5y + 7)

using

sub for x

Find Point of Intersection, and state what type of line it is in relation to the circle.

x2 + y2 = 18

y = x + 6

x2 + (x + 6)2 = 18= x2 + x2 + 12x +36 = 18= 2x2 + 12x +18 = 0= 2(x2 + 6x + 9) = 0= 2(x + 3)(x + 3) = 0

x = -3 y?y = -3 + 6 = 3

(-3 , 3)

One Point of Intersection Tangent

Key Question