Let Gamma denote a curve y = y(x) which is in | vedclass

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(B) The equation of the tangent to the curve $\Gamma$ at point $P(x, y)$ is $Y-y=y^{\prime}(X-x)$.To find the intersection with the $y$-axis,set $X=0$

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$1,2$

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(B) The equation of the tangent to the curve $\Gamma$ at point $P(x, y)$ is $Y-y=y^{\prime}(X-x)$.To find the intersection with the $y$-axis,set $X=0$: $Y_p = y - xy^{\prime}$.The point $Y_p$ is $(0, y-xy^{\prime})$. The distance $PY_p$ is given as $1$,so the distance between $(x, y)$ and $(0, y-xy^{\prime})$ is $1$.$\sqrt{(x-0)^2 + (y - (y-xy^{\prime}))^2} = 1$$\sqrt{x^2 + (xy^{\prime})^2} = 1 \Rightarrow x^2 + x^2(y^{\prime})^2 = 1$$(y^{\prime})^2 = \frac{1-x^2}{x^2} \Rightarrow y^{\prime} = \pm \frac{\sqrt{1-x^2}}{x}$.Since the curve is in the first quadrant and passes through $(1,0)$,and the tangent length is constant,the slope must be negative for the curve to decrease towards the $x$-axis. Thus,$y^{\prime} = -\frac{\sqrt{1-x^2}}{x}$.This gives the differential equation $xy^{\prime} + \sqrt{1-x^2} = 0$,which matches option $(2)$.Integrating $dy = -\frac{\sqrt{1-x^2}}{x} dx$:Let $x = \sin	heta$,then $dx = \cos	heta d	heta$.$y = -\int \frac{\cos	heta}{\sin	heta} \cos	heta d	heta = -\int \frac{\cos^2	heta}{\sin	heta} d	heta = -\int \frac{1-\sin^2	heta}{\sin	heta} d	heta = \int \sin	heta d	heta - \int \csc	heta d	heta$$y = -\cos	heta - \ln|\csc	heta - \cot	heta| + C = -\sqrt{1-x^2} - \ln\left|\frac{1-\sqrt{1-x^2}}{x}ight| + C$Using $y(1)=0$,we find $C=0$. Simplifying the logarithm,we get $y = \ln\left(\frac{1+\sqrt{1-x^2}}{x}ight) - \sqrt{1-x^2}$,which matches option $(1)$.

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