A moving point remains equidistant from the point (a,0)and the line y=x. Find the equation of its locus.

Let (x1,y1) be the moving point which is equidistant from
(a,0) and  y =  x.

Then distance d of the point (x1,y1)
from, (a,0) is  d^2 = (x1-a)^2+(y1-0)^2...(1)

Therefore
 distance d of the point  (h,k) from a line ax+by+c = 0 is given
by:

d =
(ah+by+c)^2|/(a^2+b^2)

The distance d of the point from y =
x , or x-y+0 = 0 from a point (x1,y1) is given by : d^2 =
(x1-y1)^2/(1^2+1^2)^(1/2)}....(2).

Therefore from (1) and
(2)  the locus of the point (x1,y1) is given by:

(x1-a)^2
+y1^2 = (x1-y1)^2/2

2(x1^2-2ax1+a^2 + y1^2 = x1^2
-2x1y1+y1^2.

x1^2-4ax1+2a^2 +2x1y1 =
0

x1^2 +2(y1-2a)x1+ 2a^2 =
0

So by dropping the suffixes, we get the equation of the
locus.

x^2+2(y-2a)+2a^2 =
0.

Therefore the equation of the locus is 
x^2+2(y-2a)+2a^2 = 0.