If ` x^2 + 4y ^2 - 8x +12 =0` is satified by real values of x nad y then y must lies between

If the equation y^2+x^2-10x+21=0 is satisfied by real values of x and y, prove that x lies between 3 and 7 and y lies between (-2) and 2.

If 8x^2 + y^2 -12x -4xy+9 =0 , then the value of (14x-5y) is:

3x^(2) + 8x +4=0, 4y^(2)-19y + 12=0

If the circumference of the circle x^(2) + y^(2) + 8x + 8y - b = 0 is bisected by the circle x^(2) + y^(2) - 2x + 4y + a = 0 , then a + b =