Cho hai số thực x,y thỏa mãn x^2 + y^2 - 4x + 6y + 4 + căn bậc hai (y^2 + 6y + 10) = căn bậc hai(6 + 4x - x^2)

Ta có  $x^2+y^2-4x+6y+4+\sqrt{y^2+6y+10}=\sqrt{6+4x-x^2}$

$\begin{array}{l}\iff x^2+y^2-4x+6y+4+\sqrt{y^2+6y+10}-\sqrt{6+4x-x^2}=0\\ \iff x^2+y^2-4x+6y+4+\frac{\left(\sqrt{y^2+6y+10}-\sqrt{6+4x-x^2}\right)\left(\sqrt{y^2+6y+10}+\sqrt{6+4x-x^2}\right)}{\sqrt{y^2+6y+10}+\sqrt{6+4x-x^2}}=0\end{array}$

$\begin{array}{l}\iff x^2+y^2-4x+6y+4+\frac{y^2+6y+10-6-4x+x^2}{\sqrt{y^2+6y+10}+\sqrt{6+4x-x^2}}=0\\ \iff x^2+y^2-4x+6y+4+\frac{x^2+y^2-4x+6y+4}{\sqrt{y^2+6y+10}+\sqrt{6+4x-x^2}}=0\end{array}$

$\iff \left(x^2+y^2-4x+6y+4\right)\left(1+\frac{1}{\sqrt{y^2+6y+10}+\sqrt{6+4x-x^2}}\right)=0$

$1+\frac{1}{\sqrt{y^2+6y+10}+\sqrt{6+4x-x^2}}>0$

$\iff {\left(x-2\right)}^2+{\left(y+3\right)}^2=9$