Question

CHALLENGE QUESTION 6

The orthogonal trajectories of the circles passing through the points (0, \(\pm a\)) are \(x^2 + y^2 =\)

(a) \(cx - a^2\) (b) \(cx + a^2\)
(c) \(cy-a^2\) (d) \(cy + a^2\)

Answer 1

@eliassaab

@waterineyes
@satellite73

Answer 2

@mukushla
@Neemo
@UnkleRhaukus

Answer 3

@sami-21

Answer 4

x-axis is the axis of symmetry for such circles so the center lies on the x-axis

let \((b,0)\) be the centers of the circles


so the family of circles are given by: \[(x-b)^2+y^2=b^2+a^2\]

Answer 5

it gives\[x^2-2bx+y^2=a^2\]differentiate
\[2x-2b+2yy'=0\]\[x-\frac{1}{2x}(x^2+y^2-a^2)+yy'=0\]\[x^2-y^2+a^2+2xyy'=0\]so the differential equation of orthogonal trajectories is\[x^2-y^2+a^2-2xy\frac{1}{y'}=0\]\[(x^2-y^2+a^2) \ \text{d}y-2xy \ \text{d}x=0\]

Answer 6

or \(M\ dx+N\ dy=0\) where \(M=2xy\) and \(N=y^2-x^2-a^2\) \[\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\]but \[\frac{1}{N} (\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})=-\frac{2}{y}\] so the integration factor is \[e^{\int -\frac{2}{y}dy}=\frac{1}{y^2}\]

Answer 7

so diff equation becomes \[\frac{2x}{y} \ dx-\frac{x^2-y^2+a^2}{y^2}\ dy=0\]\[\frac{2x-x^2 \ dy}{y^2}+dy-\frac{a^2}{y^2}dy=0\]\[d(\frac{x^2}{y})+(1-\frac{a^2}{y^2})dy=0\] integrate \[\frac{x^2}{y}+y+\frac{a^2}{y}=c\]finally\[x^2+y^2+a^2=cy\]and\[x^2+y^2=cy-a^2\]

Answer 8

Wonderful work @mukushla deserve many kudos.................. the perfect solution for this question......... and ofcourse your answer is absolutely correct.........