Chapter 1 - First-Order Differential Equations - 1.3 The Geometry of First-Order Differential Equations - Problems - Page 32: 12

$(x-2)^2+y^2=2^2$

We are given: $$(x-c)^2+y^2=c^2$$ $$2(x-c)+2yy'=0$$ $$\frac{x-c}{y}=y'$$ Solve for c: $$\frac{x+y}{2}=c$$ Then $$y'=\frac{x-\frac{x+y}{2}}{y}$$ $$y'=\frac{y^2-x^2}{2xy}$$ Since $y(2)=2$ $$(2-c)^2+2^2=c^2$$ $$C=2$$ Therefore, the equation is equal to $(x-2)^2+y^2=2^2$