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Q. The locus of the centre of the circle of radius 3 whih rolls on outside of the...
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The locus of the centre of the circle of radius 3 units which rolls on the outside of the given circle would be another circle, concentric to the given circle, of radius 3 units larger.
First convert the given circle equation `x^2 + y^2 + 3x -6y - 9 = 0` into the standard, `(x-h)^2+(y-k)^2=r^2` form as:
`x^2 + 3x + y^2 - 6y - 9 = 0`
`rArr x^2 + 3x + 9/4 + y^2 - 6y + 9 - 9 - 9/4-9 = 0`
`rArr (x + 3/2)^2 + (y -3)^2 -81/4 = 0`
`rArr (x + 3/2)^2 + (y -3)^2 = 81/4 = (9/2)^2`
Compare this with the standard form to get,
Center of the given circle (h,k) = (-3/2,3)
radius of the given circle = r = 9/2 units
The center of the required concentric circle = (-3/2,3) and its radius = 9/2 + 3 = 15/2
Therefore, the locus of the center of the circle of radius 3 units which rolls on the outside of the given circle is:
`(x + 3/2)^2 + (y - 3)^2 = (15/2)^2`
Hence, the correct answer is option D).
Posted by llltkl on July 24, 2013 at 1:57 AM (Answer #1)
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