Question -
Answer -
(i) Centre (-2, 3) andradius 4.
Given:
The radius is 4 andthe centre (-2, 3)
By using the formula,
The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 =r2
Where, p = -2, q = 3,r = 4
Now by substitutingthe values in the above equation, we get
(x – p)2 +(y – q)2 = r2
(x – (-2))2 +(y – 3)2 = 42
(x + 2)2 +(y – 3)2 = 16
x2 +4x + 4 + y2 – 6y + 9 = 16
x2 + y2 +4x – 6y – 3 = 0
∴ The equation of thecircle is x2 + y2 + 4x – 6y – 3 = 0
(ii) Centre (a, b) andradius
.
Given:
The radius is
and the centre (a, b)
By using the formula,
The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 =r2
Where, p = a, q = b, r=
Now by substitutingthe values in the above equation, we get
(x – p)2 +(y – q)2 = r2
(x – a)2 +(y – b)2 =
x2 –2ax + a2 + y2 – 2by + b2 = a2 +b2
x2 + y2 –2ax – 2by = 0
∴ The equation of thecircle is x2 + y2 – 2ax – 2by = 0
(iii) Centre (0, -1) andradius 1.
Given:
The radius is 1 andthe centre (0, -1)
By using the formula,
The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 =r2
Where, p = 0, q = -1,r = 1
Now by substitutingthe values in the above equation, we get
(x – p)2 +(y – q)2 = r2
(x – 0)2 +(y – (-1))2 = 12
(x – 0)2 +(y + 1)2 = 1
x2 + y2 +2y + 1 = 1
x2 + y2 +2y = 0
∴ The equation of thecircle is x2 + y2 + 2y = 0.
(iv) Centre (a cos α, a sinα) and radius a.
Given:
The radius is ‘a’ andthe centre (a cos α, a sin α)
By using the formula,
The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 =r2
Where, p = a cos α, q= a sin α, r = a
Now by substitutingthe values in the above equation, we get
(x – p)2 +(y – q)2 = r2
(x – a cosα)2 +(y – a sinα)2 = a2
x2 –(2acosα)x + a2cos2α + y2 – (2asinα)y + a2sin2α= a2
We know that sin2θ+ cos2θ = 1
So,
x2 –(2acosα)x + y2 – 2asinαy + a2 = a2
x2 + y2 –(2acosα)x – (2asinα)y = 0
∴ The equation of thecircle is x2 + y2 – (2acosα) x – (2asinα) y =0.
(v) Centre (a, a) andradius √2 a.
Given:
The radius is√2 a and the centre (a, a)
By using the formula,
The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 =r2
Where, p = a, q = a, r= √2 a
Now by substitutingthe values in the above equation, we get
(x – p)2 +(y – q)2 = r2
(x – a)2 +(y – a)2 = (√2 a)2
x2 –2ax + a2 + y2 – 2ay + a2 = 2a2
x2 + y2 –2ax – 2ay = 0
∴ The equation of thecircle is x2 + y2 – 2ax – 2ay = 0.