Question

Give a step-by-step procedure for finding the equation of the osculating circle at a point on a planar curve where the curvature is nonzero.

Short answer

The step has been explained.

The equation of the osculating circle is found by the formula

$(x-h{)}^2+(y-k{)}^2=r^2$

Step-by-step solution

Step 1. Given information.

We have to give a step-by-step procedure for finding the equation of the osculating circle at a point on a planar curve where the curvature is nonzero.

Step 2. Steps to find the equation of the osculating circle at a point on a planar curve where the curvature is nonzero 

Step 1. Let $\left(x_0,y_0\right)$be a point on the curve.

Step 2. If the planar curve is of the y=f(x) then the curvature of f is determined by:

$k=\frac{\left|f^{\prime \prime }\left(x\right)\right|}{{\left(1+f^{\mathrm{\prime }}(x{)}^2\right)}^{\frac{3}{2}}}$

If the planar curve is a vector function $r\left(t\right)=⟨x\left(t\right),y\left(t\right)⟩$in the xy plane, then the curvature k of c at a point on the curve is given by :

$k=\frac{\left|x^{\mathrm{\prime }}\left(t\right)y^{\prime \prime }\left(t\right)-x^{\prime \prime }\left(t\right)y^{\mathrm{\prime }}\left(t\right)\right|}{{\left({\left(x^{\mathrm{\prime }}\left(t\right)\right)}^2+{\left(y^{\mathrm{\prime }}\left(t\right)\right)}^2\right)}^{\frac{3}{2}}}$

Step 3. The radius of the curvature is :

$\rho =\frac{1}{k}\text{if}\rho =\frac{1}{k}$

Step 4. The normal vector, N at is to be determined.

Step 5. The radius of the osculating circle is $\rho$and the centre of the osculating circle is

Step 6. The equation of the osculating circle is found by the formulalocalid="1649997995569" $(x-h{)}^2+(y-k{)}^2=r^2$