If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to 
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$

$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfies $(x-2)^2+(y+1)^2=2^2$
The expression $x^2+y^2-10x-6y+34$ can be written as $(x-5)^2+(y-3)^2$
But i do not know how to further solve it.Please help.

Two circles have at least one common point only if the distance between their centers is equal to or smaller than the sum of their radii.
All you need to do is rewrite equations as circles, extract the center, the radius of each circle and check.
$$(x-2)^2+(y+1)^2=4, a_{0}=2, b_{0}=-1, r_{0}=2$$ 
$$(x-5)^2+(y-3)^2=\frac{1}{2}, a_{1}=5, b_{1}=3, r_{1}=\frac{\sqrt{2}}{2}$$ 
$$(x-5)^2+(y-3)^2=8, a_{2}=5, b_{2}=3, r_{2}=2\sqrt{2}$$ 
$$(x-5)^2+(y-3)^2=2, a_{3}=5, b_{3}=3, r_{3}=\sqrt{2}$$
$$(x-5)^2+(y-3)^2=3, a_{4}=5, b_{4}=3, r_{4}=\sqrt{3}$$
For example you need to prove:
$$\sqrt{(5-2)^2+(3+1)^2}=5>2+\frac{\sqrt{2}}{2}$$
$$5>2+2\sqrt{2}$$
$$5>2+\sqrt{2}$$
$$5>2+\sqrt{3}$$
All of them pretty much obvious.
