<s>[INST] 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.
 [/INST] 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.
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