If ax + by = 6, bx - ay = 2 and x^2 + y^2 = 4, then the value of (a^2 + b^2) would be:

it is given that

ax + by = 6..........(1)

bx - ay = 2........(2)

and x^2 + y^2 = 4

now multiply 1 and 2nd equation by a and b respectively

we get

$$a^2$$x + $$ab$$y = 6a

$$b^2$$x - aby = 2b

adding above equations we get,

$$a^2 + b^2$$ x = 6a +2b

x = $$\frac{6a+2b}{a^2 + b^2}$$

Similarly ,

we get y = $$\frac{6a-2b}{a^2 + b^2}$$

putting above values in x^2 + y^2 = 4

we get , a = 1 and b = 3

hence $$1^2 + 3^2$$ = 10