+1904 Since it was not specifically specified, I will solve for x first and then solve for y second.
Solve for x.
\(\sqrt{(x-1)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-2)^2}=4\)
\(2\sqrt{(x-1)^2+(y-2)^2}=4\)
\((2\sqrt{(x-1)^2+(y-2)^2})^2=4^2\)
\(4((x-1)^2+(y-2)^2)=16\)
\(\frac{4((x-1)^2+(y-2)^2)}{4}=\frac{16}{4}\)
\((x-1)^2+(y-2)^2=4\)
\((x-1)^2+(y-2)^2-(y-2)^2=4-(y-2)^2\)
\((x-1)^2=4-(y-2)^2\)
\(\sqrt{(x-1)^2}=±\sqrt{4-(y-2)^2}\)
\(x-1=±\sqrt{4-(y-2)^2}\)
\(x-1=±\sqrt{4-(y^2-4y+4)}\)
\(x-1=±\sqrt{4-y^2+4y-4}\)
\(x-1=±\sqrt{-y^2+4y}\)
\(x-1=±\sqrt{y(-y+4)}\)
\(x-1=±\sqrt{y(4-y)}\)
\(x-1+1=±\sqrt{y(4-y)}+1\)
\(x-=±\sqrt{y(4-y)}+1\)
\(x=\sqrt{y(4-y)}+1,\) \(x=-\sqrt{y(4-y)}+1\)
Solve for y
\(\sqrt{(x-1)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-2)^2}=4\)
\(2\sqrt{(x-1)^2+(y-2)^2}=4\)
\((2\sqrt{(x-1)^2+(y-2)^2})^2=4^2\)
\(4((x-1)^2+(y-2)^2)=16\)
\(\frac{4((x-1)^2+(y-2)^2)}{4}=\frac{16}{4}\)
\((x-1)^2+(y-2)^2=4\)
\((x-1)^2+(y-2)^2-(x-1)^2=4-(x-1)^2\)
\((y-2)^2=4-(x-1)^2\)
\(\sqrt{(y-2)^2}=±\sqrt{4-(x-1)^2}\)
\(y-2=±\sqrt{4-(x-1)^2}\)
\(y-2=±\sqrt{4-(x^2-2x+1)}\)
\(y-2=±\sqrt{4-x^2+2x-1}\)
\(y-2=±\sqrt{3-x^2+2x}\)
\(y-2=±\sqrt{-x^2+2x+3}\)
\(y-2=±\sqrt{(-x+3)(x+1)}\)
\(y-2+2=±\sqrt{(-x+3)(x+1)}+2\)
\(y=±\sqrt{(-x+3)(x+1)}+2\)
\(y=\sqrt{(-x+3)(x+1)}+2,\) \(y=-\sqrt{(-x+3)(x+1)}+2\)
