How do you solve this geometric question using the distance forumla?

Question

I dont need an answer, I would like the way to do it using distance formula. Here is the question:

A point is at a distance 4 (square root of 2) from (-3/2, -5/2) and at a distance of 2 (square root of 5) from (9/2, -5/2). Find the point.

If you could tell me what theorem or other thing to use along with the distance forumla, It would be appreciated.

What I meant is

The square root of 32 (4 Square roots of 2) and the square root of 20 (2 square roots of 5)

so If I rephrase it it becomes this:

A point is at adistance of the square root of 32 (4 times square root of 2) from (-3/2, -5/2) and at a distance of the square root of 20 (2 times square root of 5) from (9/2, -5,2). Find the point.

Is this clearer?

Answer 1

Since when has 2 been the square root of 5?
in my world the square root of 5 is a figure that has an infinite number of digits....the first 10 being 2.236067977

and 4 is not the square root of 2 either, it`s 2 squared.

This question really hasn't been asked or explained properly.

Answer 2

The locus of points equidistant from a given point is a circle. Two circles either don't intersect at all, just touch in one point, or intersect in two points.
The equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2
Where the center is at (h, k) and the radius is r

First circle[call it ONE]:
(x + 3/2)^2 + (y + 5/2)^2 = (4√2)^2 = 32
(y + 5/2)^2 = 32 - (x - 9/2)^2

Second circle[call it TWO]:
(x - 9/2)^2 + (y + 5/2)^2 = (2√5)^2 = 20
(y + 5/2)^2 = 20 - (x - 9/2)^2

Set the right side of ONE equal to the right side of TWO:
32 - (x + 3/2)^2 = 20 - (x - 9/2)^2
12 = (x + 3/2)^2 - (x - 9/2)^2
12 = x^2 + 3x + 9/4 - (x^2 - 9x + 81/4)
12 = 12x + 9/4 - 81/4
12 = 12x -18
30 = 12x
x = 5/2
Substitute into ONE to get y:
(x + 3/2)^2 + (y + 5/2)^2 = 32
(5/2 + 3/2)^2 + (y + 5/2)^2 = 32
16 + (y + 5/2)^2 = 32
(y + 5/2)^2 = 16
y + 5/2 = +/- 4
y = 3/2 +/- 4
y = 3/2, -13/2

Answer 3

d = sqrt (xsub2 - xsub1)^2 + (ysub2 - ysub1)^2

Answer 4

W.H.A.T