Maths(Coordinate Treatment of Simple Locus Problems)

By the following formulas:
For any circle with equation: x² + y² + Dx + Ey + F =0,
centre = ( -D/2, -E/2)　＆　raidus = (1/2)√(D² + E² - 4F)
distance between two points
= √[(difference between x-coordinates)² + (difference between y-coordinates)²]



1. Consider the circle C: x² + y² - 2x + 4y - 5 =0
(a.) centre = ( -(-2)/2, -4/2) = (1, -2)
radius = (1/2)√[(-2)² + 4² - 4(-5)] = (1/2)[√(40)] =√(10)


(b.)
Method 1: Substit ute A(2,-4) into x² + y² - 2x + 4y - 5
2² + (-4)² - 2(2) + 4(-4) - 5
= -5 ( < 0 )
therefore, A(2,-4) lies inside circle C
Method 2: distance between A(2,-4) and centre of C (1, -2)
√{(2-1)²] + [(-4)-(-2)]²} =√5 ( < radius of C: √(10) )
therefore, A(2,-4) lies inside circle C
(c.)
Method 1: Substitute B(-3, 1) into x² + y² - 2x + 4y - 5
(-3)² + (1)² - 2(-3) + 4(1) - 5
=15 ( > o)
therefore, B(-3, 1) lies outside circle C



Method 2: distance between B(-3, 1) and centre of C (1, -2)
√{(-3-1)²] + [1-(-2)]²} =5 ( > radius of C: √(10) )
therefore, B(-3, 1) lies outside circle C


2.
(a)
For circles C1: x²+y²+2x+4y-3=0
centre = ( -2/2, -4/2) = (-1, -2)
radius = (1/2)√[2² + 4² - 4(-3)] = (1/2)[√(32)] =2√2


For circle C2: X²+Y² - 8X - 6Y+7=0
centre = ( -(-8)/2, -(-6)/2) = (4, 3)
radius = (1/2)√[(-8)² + (-6)² - 4(7)] = (1/2)[√(72)] =3√2


(b)
distance between centre of C1(-1, -2) & centre of C2 (4,3)
= √[(-1-4)² + (-2-3)²] =√(50) = 5√2
radius of C1 + radius of C2 = 2√2 + 2√2 = 5√2


As distance between centre of C1 & C2 = radius of C1 + radius of C2
therefore, the two circles touch each other .

1.Consider the circle C:x^2+y^2-2x+4y-5=0.
(a)Find the centre and the redius of circle C.
Centre:[-2/-2 , 4/-2] = (1,-2)
Radius:(1/2){[2^2 + 4^2 - 4(-5)]^1/2} = (10)^1/2
(b)Show that A(2,-4) lies inside circle C.
the distance of this point to centre:
[(2 - 1)^2 + (-4 + 2)^2]^1/2
= 5^1/2
<10^1/2
<radius
it lies inside.

(c)Show that B(-3,1) lies outside circle C.
the distance of this point to centre:
[(1 + 3)^2 + (-2 - 1)^2]^1/2
= 5
>10^1/2
>radius
it lies outside.


2.Consider two circlesC1:x^2+y^2+2x+4y-3=0 and C2:X^2+y^2-8X-6Y+7=0.
(a)Find the centres and the radii of C1 and C2.
For C1,
centre:(-1,-2),radius:(1/2)[2^2 + 4^2 - 4(-3)]^1/2 = 2(2^1/2)
For C2:
centre:(4,3),radius:(1/2)[8^2 + 6^2 - 4(7)]^1/2 = 3(2^1/2)

(b)Find the distance vetween the two centres.
Distance:
[(4 + 1)^2 + (3 + 2)^2]^1/2
= 5[(2)^1/2]

(c)Hence show that the two circles touch each other.
the radius of C1 + the radius of C2
= 3(2)^1/2 + 2(2)^1/2
= 5(2^1/2)
= distance between two centre,
so,C1,C2 touches each other.