Question

Find the differential equation of families of curve:
All circles. Use the curvature

Answer 1

first write the general eq. of circle .
Let (h,k) be the centre and r be the radius of the circle
eq. of circle is\[\left( x-h \right)^{2}+\left( y-k \right)^{2}=r ^{2}\]
diff. twice w.r.t x to eliminate a & b
\[2\left( x-h \right)+2\left( y-k \right)y'=0\]
\[or \left( x-h\right)+\left( y-k \right)y'=0\]
\[or \left( x-h \right)=-\left( y-k \right)y'....(1)\]
again diff.
1=-[y'.y'+((y-k)y"]
\[or 1=-[\left( y' \right)^{2}+\left( y-k \right)y"]\]
\[\left( y-k \right)y"=-\left( 1+\left( y' \right)^{2}\right)\]
\[(y-k)=-\frac{ 1+\left( y' \right)^{2} }{ y" }\]

Answer 2

from (1) \[ x-h=+\frac{ 1+\left( y' \right) ^{2}}{y" }y'...(2)\]

Answer 3

plug the values of x-h and y-k in
\[\left( x-h \right)^{2}+\left( y-k \right)^{2}=r ^{2}\]
\[solve and put y'=\frac{ dy }{dx },y"=\frac{ d ^{2}y }{ dx ^{2} }\]

Answer 4

i am not sure you want like this or something else.

Answer 5

can u give the final answer?

Answer 6

\[\left[ \frac{ 1+\left( y' \right)^{2} }{y"} \right]^{2}\left( y' \right)^{2}+\left[ \frac{ 1+\left( y' \right)^{2} }{ y" } \right]^{2} =r ^{2} \]
\[\left\{ 1+\left( y' \right)^{2} \right\}^{2}\left\{ \left( y' \right)^{2} +1\right\}=r ^{2 }\left( y" \right)^{2}\]
\[\left[ 1+\left( y' \right)^{2} \right]^{3}=r ^{2}\left( y" \right)^{2}\]
\[\left[ 1+\left( \frac{ dy }{ dx } \right)^{2} \right]^{3}=r ^{2}\left( \frac{ d ^{2} y}{dx ^{2} } \right)^{2}\]