Proving that the intersection of $(a-4)^{2}+(b-1)^{2}<4$ and $(a-1)^{2}+(b-5)^{2}<4$ is empty I´m trying to prove that these two inequalities intersection is empty, but i don´t know how to proceed i know they are circles, but i´m trying to do it with only algebra
Let $a,b\in\mathbb{R}$ and the inequalities are
$$(a-4)^{2}+(b-1)^{2}<4 \quad\land\quad (a-1)^{2}+(b-5)^{2}<4$$
Any insight would be appreciated.
 A: Lets suppose that $(a,b)$ belongs to the set of points where
$$ \sqrt{(a-1)^2+(b-5)^2} <2. $$
By the triangle inequality for vectors in $\mathbb R^2$, in the form $|x|\ge |x-y|-|y| $, where $|x|:=\sqrt{x_1^2+x_2^2}$,
$$\sqrt{(a-4)^2+(b-1)^2} \ge \sqrt{(4-1)^2+(1-5)^2}-\sqrt{(a-1)^2+(b-5)^2}   > 5 - 2 =3 $$
which implies that $(a,b)$ cannot belong to the set of points where
$$ \sqrt{(a-4)^2+(b-1)^2} < 2.$$ So there are no points in the intersection.
A: The distance between the centers of the two circles $= \sqrt{(4-1)^2 + (1-5)^2} = 5$
is less than the sum of the radii of the two circles $= 2 + 2 = 4$
A: Expand both equations to get: $$a^2-8a+16+b^2-2b+1 < 4$$ $$a^2-2a+1+b^2-10b+25 < 4$$ Add these two together and divide by $2$ to get: $$a^2-5a+b^2-6b+\dfrac{43}{2} < 4.$$ Now, we use the "complete the square" trick on the left side twice: $$a^2-5a+\dfrac{25}{4}+b^2-6b+9+\dfrac{43}{2} < 4+\dfrac{25}{4}+9$$ $$\left(a-\dfrac{5}{2}\right)^2+(b-3)^2+\dfrac{43}{2} < \dfrac{77}{4}$$ $$\left(a-\dfrac{5}{2}\right)^2+(b-3)^2 < -\dfrac{9}{4}$$ This is clearly a contradiction, since the LHS must be positive. Therefore, there are no reals $(a,b)$ which satisfy both inequalities.
A: Put A = (4,1), B = (1,5), C = (a,b). Then by property of triangle ABC : 5 = AB < CA + CB < 2+ 2 = 4. Thus : 5 < 4 : contradiction. Thus you can’t have both of inequalities above to hold .
A: 
i know they are circles,

And that's key.
The are the interior of circles (not actually circles themselves) with radius $2$.  And centered and $(4,1)$ and $(1,5)$. If the distance between the centers is more than sum of the radii the circles must be disjoint.
And indeed $||(4, 1),(1,5)|| = \sqrt{(4-1)^2 + (1-5)^2}=\sqrt{3^2 + (-4)^2}= \sqrt{9+16}=\sqrt{25} = 5 > 2+ 2$
And that proves the circles are disjoint.
You can prove this directly.  If $(a-4)^2 +(b-1)^2 < 4$ Then as squares are non negative we must have $(b-1)^2 < 4$ so $|b-1| < 2$ and $-1 < b < 3$.
This means that $-6 < b-5 <-2$ and therefore $36 > (b-5)^2 > 4$.
But that means $(a-1)^2 + (b-5)^2 > 4$.
So you can't have both $(a-4)^2 +(b-1)^2 < 4$ and  $(a-1)^2 + (b-5)^2< 4$
