Let [dy dx = ax - by + a bx + cy + a,,a,b,c R], represents a circle with center ([], []).

Question

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 :

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 $$\beta  = 1$$

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

Solutions · 1