The differential equation of the family of ci | vedclass

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(C) Let the center of the circle be $(h, h)$ since it lies on the line $y=x$. Since the circle passes through the origin $(0,0)$,its radius $r$ is the

Vedclass · Accepted Answer

$(x^2-y^2+2xy) dx = (x^2-y^2-2xy) dy$

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(C) Let the center of the circle be $(h, h)$ since it lies on the line $y=x$. Since the circle passes through the origin $(0,0)$,its radius $r$ is the distance from $(h, h)$ to $(0,0)$,so $r^2 = h^2 + h^2 = 2h^2$. The equation of the circle is $(x-h)^2 + (y-h)^2 = 2h^2$. Expanding this,we get $x^2 - 2xh + h^2 + y^2 - 2yh + h^2 = 2h^2$,which simplifies to $x^2 + y^2 - 2h(x+y) = 0$. Differentiating with respect to $x$,we get $2x + 2yy' - 2h(1+y') = 0$,which gives $h = \frac{x+yy'}{1+y'}$. Substituting $h$ back into the circle equation: $x^2 + y^2 = 2(\frac{x+yy'}{1+y'})(x+y)$. $(x^2+y^2)(1+y') = 2(x+y)(x+yy')$. $(x^2+y^2) + (x^2+y^2)y' = 2(x^2 + xyy' + xy + y^2y')$. $(x^2+y^2) + (x^2+y^2)y' = 2x^2 + 2xyy' + 2xy + 2y^2y'$. Rearranging terms to isolate $y'$: $(x^2+y^2-2xy-2y^2)y' = 2x^2 + 2xy - x^2 - y^2$. $(x^2-y^2-2xy)y' = x^2-y^2+2xy$. Thus,$(x^2-y^2+2xy) dx = (x^2-y^2-2xy) dy$.

The differential equation of the family of circles passing through the origin and having their center on the line $y=x$ is:

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