The acute angle between the curves x^2+y^2=x+ | vedclass

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(D) The given curves are $x^2+y^2-x-y=0$ $(i)$ and $x^2+y^2-2y=0$ $(ii)$.To find the intersection points,subtract $(i)$ from $(ii)$: $(x^2+y^2-2y) - (

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$\frac{\pi}{4}$

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(D) The given curves are $x^2+y^2-x-y=0$ $(i)$ and $x^2+y^2-2y=0$ $(ii)$.To find the intersection points,subtract $(i)$ from $(ii)$: $(x^2+y^2-2y) - (x^2+y^2-x-y) = 0 \Rightarrow x-y=0 \Rightarrow x=y$.Substitute $x=y$ into $(ii)$: $x^2+x^2=2x \Rightarrow 2x^2-2x=0 \Rightarrow 2x(x-1)=0$. Thus,the intersection points are $(0,0)$ and $(1,1)$.For curve $(i)$,differentiating w.r.t. $x$: $2x+2y\frac{dy}{dx}=1+\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1-2x}{2y-1} = m_1$.For curve $(ii)$,differentiating w.r.t. $x$: $2x+2y\frac{dy}{dx}=2\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{1-y} = m_2$.At $(1,1)$,$m_1 = \frac{1-2}{2-1} = -1$ and $m_2 = \frac{1}{1-1}$ (undefined,vertical tangent).Since one tangent is vertical,the angle $	heta$ is given by $|	an 	heta| = |\frac{1}{m_1}| = |\frac{1}{-1}| = 1$.Therefore,$	heta = \frac{\pi}{4}$.

The acute angle between the curves $x^2+y^2=x+y$ and $x^2+y^2=2y$ is

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