Find the coordinates of the point which is at a distance of 2 units from (5,4) and 10 units from (11, - 2).?

Let (a, b) be the coordinates of the point.

The distance of (a, b) and (5, 4) is 2 :
√[(a - 5)² + (b - 4)²] = 2
{√[(a - 5)² + (b - 4)²]}² = 2²
(a - 5)² + (b - 4)² = 4
a² - 10a + 25 + b² - 8b + 16 = 4
a² + b² - 10a - 8b + 37 = 0 ...... [1]

The distance of (a, b) and (11, -2) is 10 :
√[(a - 11)² + (b + 2)²] = 10
{√[(a - 11)² + (b + 2)²]}² = 10²
(a - 11)² + (b + 2)² = 100
a² - 22a + 121 + b² + 4b + 4 = 100
a² + b² - 22a + 4b + 25 = 0 ...... [2]

[1] - [2] :
12a - 12b + 12 = 0
a - b + 1 = 0
b = a + 1 ...... [3]

Substitute [3] into [1] :
a² + (a + 1)² - 10a - 8(a + 1) + 37 = 0
a² + a² + 2a + 1 - 10a - 8a - 8 + 37 = 0
2a² - 16a + 30 = 0
a² - 8a + 15 = 0
(a - 3)(a - 5) = 0
a = 3 or a = 5

When a = 3 :
Substitute a = 3 into [3] :
b = 3 + 1
b = 4

When a = 5 :
Substitute a = 5 into [3] :
b = 5 + 1
b = 6

The point is (3, 4) or (5, 6).

Some 'heavy' algebra here, but here goes!!!!

(x - 5)^2 + ( y - 4)^2 = 2^2
(x - 11)^2 + (y + 2)^2 = 10^2

x^2 - 10x + 25 + y^2 - 8y + 16 = 4
x^2 - 22x + 121 + y^2 + 4y + 4 = 100

x^2 + y^2 -10x - 8y = -37
x^2 + y^2 -22x + 4y = -25

Subtract
12x -12y = -12
x - y = -1 ( factoring '12')
If x = 0 then y = 1
or if x = -1 then y = 0

So there are two possibilities ( x,y) = ( 0,1) & ( -1, 0)