Question

Find the equivalent D.E. by eliminating the arbitrary constant: (x-a)^2+(y-b)^2=1

Answer 1

unit circles at (a,b)

Answer 2

how?

Answer 3

\[(x-a)^2+(y-b)^2=1\]\[(y-b)^2=1-(x-a)^2\]
\[\frac{\text d}{\text dx}\left((y-b)^2\right)=\frac{\text d}{\text dx}\left(1-(x-a)^2\right)\]\[2(y-b)\frac{\text dy}{\text dx}=-2(x-a)\]\[\frac{\text dy}{\text dx}=\frac{-2(x-a)}{2(y-b)}=\frac{a-x}{y-b}\]

Answer 4

yes that's the first derivative. What's next? Thanks

Answer 5

\[\frac{\text dy}{\text dx}=\frac{a-x}{y-b}\]
is a first order separable differential equation

Answer 6

To solve the DE

separate the variables
\[({y-b})\cdot{\text dy}=({a-x})\cdot{\text dx}\] and integrate\[\int({y-b})\cdot{\text dy}=\int({a-x})\cdot{\text dx}\]\[\frac{y^2}2-by=ax-\frac {x^2} 2+c\]\[
y^2-2by=2ax-x^2+c\]\[(y-b)^2-b^2=-(a-x)^2+a^2+c\]\[(y-b)^2+(a-x)^2=a^2+b^2+c\]
\[a^2+b^2+c=1\]