Maths Line&Circle

1.
(a)
C1: x² + y² - 4x + 2y - (1/2) = 0
Centre of C1 = (4/2, -2/2) = (2, -1)

(b)
Centre of C2 = (2,-1)

Since C2 touches the x-axis, then radius of C2 = 1

C2 : (x - 2)² + (y + 1)² = 1²
C2 : x² + y² - 4x + 2y + 4 = 0


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2.
(a)
Let A = (a, 0), then B = (0, a) and centre of C = (a, a)

Radius of the circle :
√[(a - 8)² + (a - 4)²] = a
(a - 8)² + (a - 4)² = a²
a² - 24a + 80 = 0
(a - 4)(a - 20) = 0
a = 4 or a = 20

Hence, A(4,0) and B(0, 4) or A(20, 0) and B(0, 20)

(b)
When a = 4 :
C : (x - 4)² + (y - 4)² = 4²
C : x²+ y² - 8x - 8y + 16 = 0

When a = 20 :
C : (x - 20)² + (y - 20)² = 20²
C : x²+ y² - 40x - 40y + 400 = 0

(c)
When a = 4 :
Slope of AP = (4 - 0)/(8 - 4) = 1
Slope of AB = (0 - 4)/(4 - 0) = -1
Since the product of their slopes is -1, then AP⊥AB when a = 4

Diameter of the circle = 4 x 2 = 8


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3.
(a)
Slope of PR = (0 - 4)/(a - 2) = -4/(a - 2)
Slope of QR = (0 + 6)/(a - 4) = 6/(a - 4)

(b)
Since PR⊥QR, then the product oftheir slope is -1.
[-4/(a - 2)] x [6/(a - 4)] = -1
(a - 2)(a - 4) = 24
a² - 6a - 16 = 0
(a - 8)(a + 2) = 0
a =8 or a =-2

(c)
since P, Q and R lie on the circle and ∠PRQ = 90°,
then the PQ is the diameter.

Centre of C = ((2 + 4)/2, (4 - 6)/2)) = (3, -1)
Radius of C = √[(3 - 2)² + (-1 - 4)²] = √26

C : (x - 3)² + (y + 1)² = (√26)²
C : x²+ y² - 6x + 2y - 16 = 0

(d)
Distance between the origin and centre of C
= √[(3 - 0)² + (-1 - 0)²
= √10 < radius of the C

Hence, theorigin lies inside the circle C.