If a point p(x,y) is situated on a circle whose centres is (3,-2) and whose radius is 3 unit then prove that x^2+y^2-6x+4y+4=0
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Equation of circle when centre (g,f) and radius a is given :
(x - g)2 + (y - f)2 = a2
Where g = 3,f = -2, a = 3
Putting values in equation, we have
(x - 3)2 + (y + 2)2 = 32
x2- 6x + 9 + y2 + 4y + 4 = 9
x2 + y2 - 6x + 4y + 13 = 9
x2 + y2- 6x + 4y + 13 - 9 = 0
x2 + y2 - 6x + 4y + 4 = 0
Hence proved
Step-by-step explanation:
Answered by
1
Answer:
Equation of circle when centre (g,f) and radius a is given :
(x - g)2 + (y - f)2 = a2
Where g = 3,f = -2, a = 3
Putting values in equation, we have
(x - 3)2 + (y + 2)2 = 32
x2- 6x + 9 + y2 + 4y + 4 = 9
x2 + y2 - 6x + 4y + 13 = 9
x2 + y2- 6x + 4y + 13 - 9 = 0
x2 + y2 - 6x + 4y + 4 = 0
Hence proved
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