Determine the curves in the plane such that the normal at…
Determine the curves in the plane such that the normal at every point passes through one fixed point \((a,b)\).
Explanation of Method
Write the normal line at \((x,y)\), impose that it passes through the fixed point \((a,b)\), and solve the resulting separable ODE. Both the derivation of the ODE and the final solution are checked independently with Sympy — including solving the ODE directly with \(\texttt{dsolve}\) and confirming, via implicit differentiation of the candidate circle family, that the normal condition holds identically. This is the classical (and well-known) fact that the normals of a circle all pass through its center, so the result was expected to be correct; the check below confirms there is no algebraic slip in the specific derivation.
1. The differential equation
The normal at \((x,y)\) has slope \(-1/y'\); solving \(b-y=-\tfrac1{y'}(a-x)\) for \(y'\) with Sympy gives \(y'=\dfrac{x-a}{b-y}\) exactly, matching the source.
2. Solve directly with Sympy's \(\texttt{dsolve}\)
Calling \(\texttt{dsolve}\) on \(Y'(x)=\dfrac{x-a}{b-Y(x)}\) returns \(Y(x)=b\pm\sqrt{C_1+2ax+b^2-x^2}\), which rearranges to \((x-a)^2+(Y-b)^2 = C_1+a^2+b^2\) — a circle centered at \((a,b)\), confirming the source's separation-of-variables solution \((x-a)^2+(y-b)^2=r^2\) with \(r^2=C_1+a^2+b^2\).
3. Verify the property on the candidate circle
Implicitly differentiating \((x-a)^2+(y-b)^2=r^2\) in Sympy gives \(y'=\dfrac{a-x}{y-b}=-\dfrac{x-a}{y-b}\), which matches the ODE of Step 1 exactly (\(\texttt{simplify}\) of the difference is \(0\)). Substituting this slope into the normal-line equation at \((x,y)\) and evaluating at \((X,Y)=(a,b)\), Sympy simplifies the residual to \(0\) identically, confirming the normal always passes through \((a,b)\).
Result
Final answer:
The source's derivation and geometric conclusion are confirmed correct in full. The curves are exactly the circles centered at the fixed point: \[ \boxed{\ (x-a)^2+(y-b)^2=r^2\ } \] with arbitrary radius \(r\) — the normal to a circle at any point is the radius through that point, so it necessarily passes through the center.