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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
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