Question -
Answer -
(i) x2 + y2 +6x – 8y – 24 = 0
Given:
The equation of thecircle is x2 + y2 + 6x – 8y – 24 = 0 …… (1)
We know that for acircle x2 + y2 + 2ax + 2by + c = 0 …… (2)
Centre = (-a, -b)
So by comparing equation(1) and (2)
Centre = (-6/2,-(-8)/2)
= (-3, 4)
Radius = √(a2 + b2 –c)
= √(32 + 42 –(-24))
= √(9 + 16 + 24)
= √(49)
= 7
∴ The centre ofthe circle is (-3, 4) and the radius is 7.
(ii) 2x2 +2y2 – 3x + 5y = 7
Given:
The equation of thecircle is 2x2 + 2y2 – 3x + 5y = 7 (divide by 2we get)
x2 + y2 –3x/2 + 5y/2 = 7/2
Now, by comparing withthe equation x2 + y2 + 2ax + 2by + c = 0
Centre = (-a, -b)
Given:
The equation of thecircle is
(Multiply by 2 we get)
x2 + y2 +2x cos θ + 2y sin θ – 8 = 0
By comparing with theequation x2 + y2 + 2ax + 2by + c = 0
Centre = (-a, -b)
= [(-2cos θ)/2 ,(-2sin θ)/2]
= (-cos θ, -sin θ)
Radius = √(a2 + b2 –c)
= √[(-cos θ)2 +(sin θ)2 – (-8)]
= √[cos2 θ+ sin2 θ + 8]
= √[1 + 8]
= √[9]
= 3
∴ The centre andradius of the circle is (-cos θ, -sin θ) and 3.
(iv) x2 + y2 –ax – by = 0
Given:
Equation of the circleis x2 + y2 – ax – by = 0
By comparing with theequation x2 + y2 + 2ax + 2by + c = 0
Centre = (-a, -b)
= (-(-a)/2, -(-b)/2)
= (a/2, b/2)
Radius = √(a2 + b2 –c)
= √[(a/2)2 +(b/2)2]
= √[(a2/4 + b2/4)]
= √[(a2 + b2)/4]
= [√(a2 + b2)]/2
∴ The centre andradius of the circle is (a/2, b/2) and [√(a2 +b2)]/2