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.

As you note, the restriction is
$$(x-2)^2+(y+1)^2=2^2$$
That means the distance from point $P(x,y)$ to the point $A(2,-1)$, $AP$, is $2$.
As you also note, the function is
$$f(x,y)=(x-5)^2+(y-3)^2$$
which is the square of the distance of point $P(x,y)$ to point $B(5,3)$, $PB$. Note that the distance from point $A(2,-1)$ to point $B(5,3)$, $AB$, is
$$AB=\sqrt{(5-2)^2+(3--1)^2}=5$$
To find the limits on that distance we use the triangle inequalities
$$|AB-AP| \le PB \le AB+AP$$
or
$$|5-2| \le PB \le 5+2$$
$$3 \le PB \le 7$$
So the square of distance PB, or function value, ranges between $9$ and $49$. None of the multiple choices are in that range, so none of the given choices can be correct.
