Question

Show that general equation of a circle which passes through the points $(x_1,y_1)$ and $(x_2,y_2)$ may be written as $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+\lambda \left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$
and hence deduce the diameter form of the equation of a circle.

Answer 1

The first part represents a circle $S=0$ on $P(x_1,y_1)$ and $Q(x_2,y_2)$ as diameter. The second part i.e. determinant is the equation of a straight line $a=0$ which passes through the points $P$ and $Q$. The given equation is now of the form $S+\lambda u=0$ which represents a circle passing through the intersection of $S=0$ and $u=0$ i,e, the points $P$ and $Q$. In case $PQ$ is a diameter then the centre $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ will lie on the line $PQ$ and hence the determinant becomes $\left|\begin{array}{ccc}\frac{x_1+x_2}{2}& \frac{y_1+y_2}{2}& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|$ becomes zero, because $\frac{1}{2}(R_2+R_3)$ becomes identical with $R_1$. The equation $S+\lambda u=0$ in this case reduces to $S=0$ which therefore represents the circle described on $PQ$ as diameter.