# Find a circumference with center on a line

I have a set of circumferences $$x^2 + y^2 + \alpha_1 x + \beta_1 y + \gamma_1 + k(x^2 + y^2 + \alpha_2 x + \beta_2 y + \gamma_2) = 0$$ $\alpha_1, \alpha_1, \beta_1, \beta_2, \gamma_1, \gamma_2$ given.

I need to find the value of $k$ that correspond to a circumference with center on the line $$y = mx$$ I tried to set: $x_C = - \frac{\alpha_1 + k\alpha_2}{2}$ and $y_C = - \frac{\beta_1 + k\beta_2}{2}$ and the solve in $k$: $$y_C = m x_C$$

Am i doing it right?

Thank you, regards.

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yes! that is correct. – Tomarinator Apr 21 '12 at 17:21

Your strategy is correct but you should take $$x_C=-\frac{\alpha_1+k\alpha_2}{2(1+k)},\ y_C=-\frac{\beta_1+k\beta_2}{2(1+k)}.$$
Infact a circumference $x^2+y^2+ax+by+c=0,$ (where $a^2+b^2-4c>0$) has its center in $$(x_C,y_C)=(-\frac{a}{2},-\frac{b}{2}),$$ because, by completing the squares, it can be written as $(x+{a}{2})^2+(y+\frac{b}{2})^2=\frac{1}{4}(a^2+b^2-4c).$
In your specific situation the circumference is given by $$x^2+y^2+\frac{\alpha_1+k\alpha_2}{1+k}x+\frac{\beta_1+k\beta_2}{1+k}y+\textrm{const.}=0$$