find centre $ radius of the circle 4(x2+y2)+ 12ax - 6ay -a2 =0
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The general equation of circle is ;
x^2 + y^2 +2gx +2fy +c = 0
In this equation :
4x^2 + 4y^2 +12ax -6ay -a^2 =0
x^2 + y^2 + 3ax - (3/2)ay -(a^2/4) =0
g = (3/2)a
f = -(3/4)a
c = -(a^2/4)
Centre of the circle = (-g,-f)
= ( -3/2 a, 3/4 a)
Radius = √g^2+f^2+c
= √(9/4)a^2 + (9/16)a^2 + (a^2/4)
= √(49/16)a^2
= 7/4 a
hope this helped you,
plzzz mark as brainliest .........
x^2 + y^2 +2gx +2fy +c = 0
In this equation :
4x^2 + 4y^2 +12ax -6ay -a^2 =0
x^2 + y^2 + 3ax - (3/2)ay -(a^2/4) =0
g = (3/2)a
f = -(3/4)a
c = -(a^2/4)
Centre of the circle = (-g,-f)
= ( -3/2 a, 3/4 a)
Radius = √g^2+f^2+c
= √(9/4)a^2 + (9/16)a^2 + (a^2/4)
= √(49/16)a^2
= 7/4 a
hope this helped you,
plzzz mark as brainliest .........
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