Question

Find the equation to the circle whose centre is at the point $(\alpha ,\beta )$ and which passes through the origin, and prove that the equation of the tangent at the origin is $\alpha x+\beta y=0$.

Answer 1

Given the centre is at $(\alpha ,\beta )$

The equation of circle be,

$(x-\alpha {)}^2+(y-\beta {)}^2=r^2$

The circle passes through origin

$\therefore {\alpha }^2+{\beta }^2=r^2$

Hence the equation of circle becomes

$x^2+y^2-2\alpha x-2\beta y=0$

The equation of tangent at point $(x_1,y_1)$ for a standard equation of circle $x^2+y^2+2gx+2fy+c=0$ is

$x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c=0$

$gx+fy+c=0$

$g=-\alpha ,f=-\beta ,c=0$

$\therefore \alpha x+\beta y=0$