Family of Curves Try and obtain the differential equations of the given families of curves All circles passing through the origin and (0,4).

Name: Family of Curves Try and obtain the differential equations of the given families of curves All circles passing through the origin and (0,4).
Uploaded: 2022-02-23T17:22:57+00:00
Description: Family of Curves Try and obtain the differential equations of the given families of curves All circles passing through the origin and (0,4).

Question

Family of Curves  Try and obtain the differential equations of the given families of curves  All circles passing through the origin and (0,4).

puhnut1m · Accepted Answer

We have to find the differential equation of family of curve.Equation of circle is:(x−h)2+(y−k)2=r2 (1)9t passes through (0,0)So, eq (1) becomesh2+k2=r2⇒(x−h)2+(y−k)2=h2+k2 (2)Equation (1) also pass through (0,4)So, h2+(4−k)2=r2¬{h2}+(4−k)2=¬{h2}+k2(4−k)2=k24−k=k4=2kk=2So, equation (2) becomes(x−h)2+(y−2)2=h2+4x2+¬{h2}−2xh+y2+¬{4}−4y=¬{h2}+¬{4}x2−2xh+y2−4y=0 (3)Differentiating w.r.t. x,2x−2h+2ydydx−4dydx=0h=x+dydx(y−2) (4)Put in (3),x2−2x2−2xydydx+4xdydx+y2−4y=0y′(4x−2xy)−x2+y2−4y=0

Maricela Alarcon · Accepted Answer

All circles passing through the origin and (0, 4).   Let the equation of the circle be (x−h)2+(y−k)2=r2.  Now the circle passes through the origin, therefore we get  (0−h)2+(0−k)2=r2  ⇒h2+k2=r2→(1)  And also the circle passes through the point (0, 4), therefore we get  (0−h)2+(4−k)2=r2  ⇒h2+16−8k+k2=r2  ⇒h2+k2+16−8k=r2  ⇒r2+16−8k=r2 [By using equation (1)]  ⇒16−8k=0  ⇒8k=16  ⇒k=2  Now put k=2 in equation (1), we get  h2+22=r2  RIghtarrowr2=h2+4  Therefore equation of the circle becomes,  (x−h)2+(y−2)2=h2+4  ⇒x2−2hx+h2+y2−4y+4=h2+4  ⇒x2−2hx+y2−4y=0  Now differentiating with respect to x, we get  2x−2h(1)+2y⋅dydx−4⋅dydx=0  ⇒dydx(2y−4)=2h−2x  ⇒dydx=2h−2x2y−4  ⇒dydx=x2+y2−4yx−2x2y−4

karton · Accepted Answer

(x−h)2+(y−k)2=r2 (1)h2+k2=r2⇒(x−h)2+(y−k)2=h2+k2 (2)h2+(4−k)2=r2⧸h2+(4−k)2=⧸h2+k2(4−k)2=k24−k=k4=2kk=2(x−h)2+(y−2)2=h2+4x2+⧸h2−2xh+y2+⧸4−4y=⧸h2+⧸4x2−2xh+y2−4y=0 (3)2x−2h+2ydydx−4dydx=0h=x+dydx(y−2) (4)x2−2x2−2xydydx+4xdydx+y2−4y=0y′(4x−2xy)−x2+y2−4y=0

Family of Curves Try and obtain the differential equations of the

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