91Ó°ÊÓ

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 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:

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

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

$k=\frac{\left|x^{\mathrm{′}}\left(t\right)y^{′â\P IJ}\left(t\right)−x^{′â\P IJ}\left(t\right)y^{\mathrm{′}}\left(t\right)\right|}{{\left({\left(x^{\mathrm{′}}\left(t\right)\right)}^2+{\left(y^{\mathrm{′}}\left(t\right)\right)}^2\right)}^{\frac{3}{2}}}$

Step 3. The radius of the curvature is :

$\mathrm{ÒÏ}=\frac{1}{k}\text{if}\mathrm{ÒÏ}=\frac{1}{k}$

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

Step 5. The radius of the osculating circle is $\mathrm{ÒÏ}$and the centre of the osculating circle is

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