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.
 A: Since you have simplified the given equation as $(x-2)^2+(y+1)^2=2^2$, then taking the parameter $\theta$, put $x=2+2\cos\theta$ and $y=-1+2\sin\theta$.
As $\theta$ varies, you can clearly see that the point $(x,y)=(2+2\cos\theta,-1+2\sin\theta)$ is on the circle $(x-2)^2+(y+1)^2=2^2.$
Now use this parametrization in $x^2+y^2-10x-6y+34$.
So we start as,
$(2+2\cos\theta)^2+(-1+2\sin\theta)^2-10(2+2\cos\theta)-6(-1+2\sin\theta)+34$
$=29-(12\cos\theta+16\sin\theta)$
Now use the result $-\sqrt {a^2+b^2} \le a\cos\theta+b\sin\theta \le \sqrt{a^2+b^2}$.
Hence $-\sqrt{12^2+16^2} \le 12\cos\theta+16\sin\theta \le \sqrt{12^2+16^2}$
$\Rightarrow -20 \le 12\cos\theta+16\sin\theta \le 20$
$\Rightarrow 29-20 \le x^2+y^2-10-6y+34 \le 29+20$
$\Rightarrow 9 \le x^2+y^2-10x-6y+34 \le 49.$
Therefore all the given options are correct.
A: 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.
A: 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.
