Find all points on the y-axis 3 units from (-1,-2)?

Answer 1

(x + 1) ² + (y + 2) ² = 9
On y axis , x = 0:-
1 + (y + 2) ² = 9
(y + 2) ² = 8
(y + 2) = ± 2√2
y = - 2 ± 2√2
Points are ( 0 , (- 2 + 2√2) ) , ( 0 , (- 2 - 2√2))

Answer 2

My predecessor made a mistake, logical thinking without even calculating anything will tell you this: the point (-1, -2) is only 1 unit away from the y axis. So if you envision a line of length 3 that is connected to the point (-1, -2) there will be two points at which the other end of the line will hit the y axis, one above and one below the point itself. (make a drawing).

The simplest solution is to use the pythagorean theorem: Here we have two right triangles. The first one has points A=(-1, -2), B=(0, -2) and C=(0, n) where n is the point on the y axis where the distance is exactly 3. The other one is mirrored down along AB.

We know the length of one side and the hypotenuse of this triangle. Let's call the length of the unknown side a. So we have the following equation:

1^2 + a^2 = 3^2 (Pythagoras: a^2 + b^2 = c^2)
a^2 = 9 - 1 = 8
a = sqrt(8) = 2sqrt(2)

So, since -2 is vertical offset of our triangle on the y axis, -2 +2sqrt(2) and -2 - 2sqrt(2) are the points on the y axis where the distance to the point (-1, -2) is exactly 3.