Let #P≡(x,y)# be the point
Given that #P≡(x,y) # is away from #A≡(0,-3)# by #sqrt5#
and that #P≡(x,y) # is away from #B≡(3,-4)# by #sqrt5#
A and B are separated by the distance given by the relation
#(3-0)^2+(-4-(-3))^2=3^2+1^2=10#
#sqrt10=3.161#
while,
#sqrt5=2.236#
Thus,
#sqrt10<2sqrt5#
ie
By applying the distance given by the relation
#(0-x)^2+(-3-y)^2=5#
and
#(3-x)^2+(-4-y)^2=5#
Simplifying
#x^2+(3+y)^2=5#
#(3-x)^2+(4+y)^2=5#
Equating the two distances
#x^2+(y+3)^2=(x-3)^2+(y+4)^2#
Expanding
#x^2+y^2+6y+9=x^2-6x+9+y^2+8y+16#
#6y-8y+6x+9-9-16=0#
Rearanging
#6x-2y-16=0#
or
#3x-y-8=0#
The points lie on the straight line
#y=3x-8#
Substituting in #x^2+(y+3)^2=5#
#x^2+(3x-8+3)^2=5#
#x^2+(3x-5)^2=5#
#x^2+9x^2-30x+25=5#
#10x^2-30x+25-5=0#
#10x^2-30x+20=0#
#x^2-3x+2=0#
Factorising
#x^2-2x-x+2=0#
#x(x-2)-1(x-2)=0#
#(x-1)(x-2)=0#
#x-1=0# or #x-2=0#
#x=1# or #x=2#
When #x=1#
#y=3x-8#becomes
#y=3(1)-8=3-8=-5#
#(x,y)=(1,-5)#
When #x=2#
#y=3x-8#becomes
#y=3(2)-8=6-8=-2#
#(x,y)=(2,-2)#
Check:
For #P(1,-5)#
Distance from #A≡(0,-3)#to#P≡(1,-5)# is given by
#sqrt((0-1)^2+(-3-(-5))^2)#
#=sqrt((-1)^2+2^2)=sqrt(1+4)=sqrt5#
For #P(2,-2)#
Distance from #A≡(0,-3)#to#P≡(2,-2)# is given by
#sqrt((0-2)^2+(-3-(-2))^2)#
#=sqrt((-2)^2+(-1)^2)=sqrt(4+1)=sqrt5#
Hence,
the solution is
#P-=(1,-5)# and #P-=(2,-2)#
