Verified answer

let, (x,y) is the required point.

(x+1)^2+(y+1)^2=9............(1)

(x-3)^2+(y-5)^2=36............(2)

subtracting (2) from (1),

8x+12y=5

or, y=(5-8x)/12.......(3)

putting (3) in (1),

(x+1)^2+((5-8x)/12+1)^2=9

after required steps,

208x^2+16x-863=0

so, x=(-16+(16^2+4*863*208)^(1/2))/(2*208) or (-16-(16^2+4*863*208)^(1/2))/(2*208)

from graph, its absurd for x to be negative.

so, x=(-16+(16^2+4*863*208)^(1/2))/(2*208)

=1.9988204

from (3),

y=-.91588

so, (x,y)=(1.9988204,.91588)

As per your statement the distance 6, squared 36 = (x--3)^2 + (y--5)^2-->(1)

Same way the other length, 3, squared 9 = (x+1)^2 + (y+1)^2--->(2)

Here (x,y) is the unknown co ordinates of the third point.

Solving (1) and (2)

(2) multiplied by 4 makes it equal to (1)

So 4(x+1)^2 + 4(y+1)^2 = (x--3)^2 + (y--5)^2

3x^2 +3y^2 +14x +18y --26 = 0

This will be an equation of a circle, the point on which would fulfill your condition that the respective lengths are 6 and 3 units. Hence so many points!