JEE Circles Doubt

Question

Can you derive a general formula for length of intercept made by a circle $x^2+y^2+2gx+2fy+c=0$ on the line $ax+by+c=0$? If no, then is there any other method to solve these type of questions?

Answer 1

\begin{align} x^2+y^2+2gx+2fy+d &= 0 \tag{1}\label{1} ,\\ ax+by+c &= 0 \tag{2}\label{2} . \end{align}

Expression \eqref{1} represent a circle with radius $r=\sqrt{g^2+f^2-d}$ centered at $O=(-g,-f)$:

\begin{align} (x+g)^2+(y+f)^2 &=g^2+f^2-d \tag{3}\label{3} . \end{align}

Intersection points $X_+,\ X_-$ of the circle $(O,R)$ and the line through the point $B$ in direction $D=z_2-z_1$ can be found as follows:

\begin{align} X_{\pm}&=B+t_{\pm}\cdot D \tag{4}\label{4} ,\\ t_{\pm}&=\frac{-(B-O)\odot D\pm\sqrt{\left((B-O)\odot D\right)^2+|D|^2(R^2-|B-O|^2)}}{|D|^2} \tag{5}\label{5} , \end{align} where $\odot$ is the dot product.

Given \eqref{2},\eqref{3}, we can apply \eqref{4}, \eqref{5} by setting (assuming that $b\ne0$)

\begin{align} O&=(-g,-f) ,\quad z_1=(0,-\tfrac cb) ,\quad z_2=(1,-\tfrac{a+c}b) ,\quad B=z_1 ,\quad D=z_2-z_1=(1,-\tfrac ab) \tag{6}\label{6} . \end{align}

The number of intersections is defined by the discriminant \begin{align} \delta&= (a\,f-b\,g)^2-d\,(a^2+b^2)+c\,(2a\,g+2b\,f-c) \tag{7}\label{7} . \end{align}

If $\delta<0$, there are no intersections, if $\delta>0$, there are two intersection points,

\begin{align} X_{\pm}&= \left( \frac{-b\,\Big(b\,g-a\,f\pm\sqrt{\delta}\Big)-a\,c}{a^2+b^2},\ \frac{ a\,\Big(b\,g-a\,f\pm\sqrt{\delta}\Big)-b\,c}{a^2+b^2} \right) \tag{8}\label{8} , \end{align}

and if $\delta=0$, then the line is tangent to the circle at the point

\begin{align} X_+=X_-&= \left( \frac{-b\,(b\,g-a\,f)-a\,c}{a^2+b^2},\ \frac{ a\,(b\,g-a\,f)-b\,c}{a^2+b^2} \right) \tag{9}\label{9} . \end{align}

The length of intercept is just \begin{align} |X_+-X_-|&= 2\sqrt{\frac{\delta}{a^2+b^2}} \tag{10}\label{10} . \end{align}