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Find the equation of the circle......

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by likithae » Tue Aug 10, 2010 10:00 pm
If the lines 2x-3y=0 and 3x-4y=7 are two diameters of a circle of radius 7, then the equation of the circle is

(a)x^2+y^2+2x-4y-47=0
(b)x^2+y^2=49
(c)x^2+y^2-2x+2y-47=0
(d)x^2+y^2=17


help me to solve it....
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by kvcpk » Wed Aug 11, 2010 12:56 am
likithae wrote:If the lines 2x-3y=0 and 3x-4y=7 are two diameters of a circle of radius 7, then the equation of the circle is

(a)x^2+y^2+2x-4y-47=0
(b)x^2+y^2=49
(c)x^2+y^2-2x+2y-47=0
(d)x^2+y^2=17


help me to solve it....
2x-3y = 0
3x-4y = 7
9y/2 -4y = 7
y/2 = 7
y=14
x = 21

Hence centre = (21,14) and radius = 7
(x-21)^2 + (y-14)^2 = 49
x^2-42x+441 + y^2-28y+196 = 49
x^2+y^2-42x-28y+147 = 0

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by likithae » Wed Aug 11, 2010 11:30 pm
answer is option c.
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