Equation of family of circles of radius r and tangent to the y-axis:
(x ± r)^{2} + (y - k)^{2} = r^{2}

The (±) sign indicates that circles can be at the left or at the right of y-axis. There is only one arbitrary constant k, thus, the differential equation is a first degree. Note that r here is a parameter (fixed radius) and need not be eliminated.
2(x ± r) + 2(y - k)y' = 0
(x ± r) + (y - k)y' = 0
(y - k) = -(x ± r) / y'

Please double check my answer. I am answering here out from what I have remembered from this topic. It's been a long while since I encountered this type of problem.

Let r = radius of cirlce

Equation of family of circles of radius r and tangent to the y-axis:

(x ± r)

^{2}+ (y - k)^{2}= r^{2}The (±) sign indicates that circles can be at the left or at the right of y-axis. There is only one arbitrary constant k, thus, the differential equation is a first degree. Note that r here is a parameter (fixed radius) and need not be eliminated.

2(x ± r) + 2(y - k)y' = 0

(x ± r) + (y - k)y' = 0

(y - k) = -(x ± r) / y'

Thus,

$(x \pm r)^2 + \left( -\dfrac{x \pm r}{y'} \right)^2 = r^2$

$(x \pm r)^2 + \dfrac{(x \pm r)^2}{(y')^2} = r^2$

$(x \pm r)^2 (y')^2 + (x \pm r)^2 = r^2 (y')^2$

$r^2 (y')^2 - (x \pm r)^2 (y')^2 = (x \pm r)^2$

$[ \, r^2 - (x \pm r)^2 \, ] (y')^2 = (x \pm r)^2$

$\left( \dfrac{dy}{dx} \right)^2 = \dfrac{(x \pm r)^2}{r^2 - (x \pm r)^2}$

$\dfrac{dy}{dx} = \dfrac{x \pm r}{\sqrt{r^2 - (x \pm r)^2}}$

answerPlease double check my answer. I am answering here out from what I have remembered from this topic. It's been a long while since I encountered this type of problem.

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