"Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$, represents a circle with center ($\alpha$, $\beta$). Then, $\alpha$ + 2$\beta$ is equal to :"

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Accepted Answer

"

Given,

${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$

$\Rightarrow bxdy + cydy + ady = axdx - bydx + adx$

$\Rightarrow bxdy + bydx + (cy + a)dy = (ax + a)dx$

$\Rightarrow b(xdy + ydx) + (cy + a)dy = (ax + a)dx$

$\Rightarrow bd(xy) + (cy + a)dy = (ax + a)dx$

Integrating both sides, we get

$\Rightarrow b\int {d(xy) + \int {(cy + a)dy = \int {(ax + a)dx} } }$

$$ \Rightarrow b\,.\,xy + c\,.\,{{{y^2}} \over 2} + ay = {{a{x^2}} \over 2} + ax + k$$

$\Rightarrow {{a{x^2}} \over 2} - {{c{y^2}} \over 2} - bxy + ax - ay + k = 0$

For equation of circle,

Coefficient of x² = Coefficient of y²

$\therefore$ ${a \over 2} = - {c \over 2}$

$\Rightarrow a = - c$

And coefficient of $xy = 0$

$\therefore$ $- b = 0$

$\Rightarrow b = 0$

$\therefore$ Circle equation becomes,

${{a{x^2}} \over 2} + {{a{y^2}} \over 2} + ax - ay + k = 0$

$\Rightarrow {x^2} + {y^2} + 2x - 2y + {{2k} \over a} = 0$

$\therefore$ Center $= ( - g, - f) = ( - 1,1) = (\alpha ,\beta )$

$\therefore$ $\alpha = - 1$ and $\beta = 1$

$\therefore$ $\alpha + 2\beta = - 1 + 2 \times (1) = 1$

"

Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$, represents a circle with center ($\alpha$, $\beta$). Then, $\alpha$ + 2$\beta$ is equal to :

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Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$, represents a circle with center ($\alpha$, $\beta$). Then, $\alpha$ + 2$\beta$ is equal to :

Solution

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