CircleHard
Question
The locus of the centre of a circle , which touches exterbally the circle x2 + y2 - 6x - 6y + 14 = 0 abd also touches the y-axis, is given by the equation
Options
A.x2 - 6x - 10y + 14 = 0
B.x2 -10x - 6y + 14 = 0
C.y2 - 6x - 10y + 14 = 0
D.y2 - 10x - 6y + 14 = 0
Solution
Let (h, k ) be the centre of the circle which touches the circle x2 + y2 - 6x - 6y + 14 = 0 and y - axis.
The centre of given cicle is (3, 3) and radius is
Since, the circle touches y-axis, the distance from its centre to y-axis must be equal to its radius, therefore its radius is h. Again, the circles meet extermally, therefore the distance vetween two centres = sum of the radii of the two circles,
Hence, (h - 3)2 + (k - 3)2 = (2 + h)2
ie, h2 + 9 - 6h + k2 + 9 - 6k = 4 + h2 + 4h
ie, k2 - 10h + 14 = 0
Thus, the locus of (h, k) is
y2 - 10x - 6y + 14 = 0
The centre of given cicle is (3, 3) and radius is
Since, the circle touches y-axis, the distance from its centre to y-axis must be equal to its radius, therefore its radius is h. Again, the circles meet extermally, therefore the distance vetween two centres = sum of the radii of the two circles,
Hence, (h - 3)2 + (k - 3)2 = (2 + h)2
ie, h2 + 9 - 6h + k2 + 9 - 6k = 4 + h2 + 4h
ie, k2 - 10h + 14 = 0
Thus, the locus of (h, k) is
y2 - 10x - 6y + 14 = 0
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