The locus of the center of the circle of radius 2 unit which
rolls on the outside of the given circle would be a concentric circle (the
circle with a common center as that of the given circle x^2 + y^2 + 3x – 6y – 9
= 0).
The radius of the concentric circle would be 2 unit more
than the radius of the given circle (x^2 + y^2 + 3x – 6y – 9 = 0).
First convert the given circle equation x^2 + y^2 + 3x – 6y –
9 = 0 into the standard form:
X^2 + 3x + y^2 – 6y – 9 = 0
x^2 + 3x + 9/4 – 9/4 + y^2 – 6y + 9 – 9 – 9 = 0
(x + 3/2)^2 – 9/4 + (y – 3)^2 – 18 = 0
(x + 3/2)^2 + (y – 3)^2 = 9/4 + 18 =
81/4 = (9/2)^2
Comparing this to the standard form of a circle: (x – h)^2 +
(y – k)^2 = r^2
Center of the given circle (h,k) = (-3/2,3)
radius of the given circle = r = 9/2
Therefore, the center of the concentric circle = (-3/2,3)
and radius = 9/2 + 2 = 13/2
The locus (equation of the concentric circle):
(x + 3/2)^2 + (y – 3)^2 = (13/2)^2