Question

$A$ and $B$ are two points in the xy-plane, which are $2\sqrt{2}$ unit distance apart and subtend an angle of ${90}^o$ at the point $C(1,2)$ on the line $x-y+1=0$, which is larger than any angle subtended by the line segment $AB$ at any other point on the line. Find the equation(s) of the circle through the points $A,B$ and $C$.

Answer 1

As shown in the figure, equation of line is $L=x-y+1=0$

points A and B are endpoints of diameter as shown in figure. (Property of angle subtended in semicircle)

Equation of family of circles touching a line $L=0$ at point $(x_1,y_1)$ is,

${(x-x_1)}^2+{(y-y_1)}^2+\lambda L=0$

where, $\lambda$ = constant parameter

Here, $x_1=1$, $y_1=2$

$\therefore {(x-1)}^2+{(y-2)}^2+\lambda (x-y+1)=0$

$\therefore x^2-2x+1+y^2-4y+4+\lambda x-\lambda y+\lambda =0$

$\therefore x^2+y^2+x(\lambda -2)+y(-\lambda -4)+(5+\lambda )=0$

$\therefore x^2+y^2+2\left(\frac{\lambda -2}{2}\right)x+2\left(\frac{-\lambda -4}{2}\right)y+(5+\lambda )=0$

Comparing this equation with standard equation of circle i.e. $x^2+y^2+2gx+2fy+c=0$, we get,

$g=\frac{\lambda -2}{2}$

$f=\left(\frac{-\lambda -4}{2}\right)$

$c=5+\lambda$

Thus, radius of circle is given as,

$r=\sqrt{g^2+f^2-c}$

$\therefore r=\sqrt{{\left(\frac{\lambda -2}{2}\right)}^2+{(-\left(\frac{\lambda +4}{2}\right))}^2-(5+\lambda )}$

$\therefore r=\sqrt{{\left(\frac{\lambda -2}{2}\right)}^2+{\left(\frac{\lambda +4}{2}\right)}^2-5-\lambda }$

$\therefore r=\sqrt{\frac{{\lambda }^2-4\lambda +4}{4}+\frac{{\lambda }^2+8\lambda +16}{4}-5-\lambda }$

$\therefore r=\sqrt{\frac{{\lambda }^2-4\lambda +4+{\lambda }^2+8\lambda +16-20-4\lambda }{4}}$

$\therefore r=\sqrt{\frac{2{\lambda }^2}{4}}$

$\therefore r=\sqrt{\frac{{\lambda }^2}{2}}$

$\therefore r=\pm \frac{\lambda }{\sqrt{2}}$ (1)

Now, AB is diameter and $l\left(AB\right)=2\sqrt{2}$ (given)

$\therefore d=2\sqrt{2}$

$\therefore r=\sqrt{2}$ (2)

From (1) and (2),

$\therefore \sqrt{2}=\pm \frac{\lambda }{\sqrt{2}}$

$\therefore \lambda =\pm 2$

Thus, equations of circles touching line L and passing through $(1,2)$ are-

1) $(x-1)+(y-2)+2(x-y+1)=0$

$\therefore x-1+y-2+2x-2y+2=0$

$\therefore 3x-y-1=0$

2) $(x-1)+(y-2)-2(x-y+1)=0$

$x-1+y-2-2x+2y-2=0$

$\therefore -x+3y-5=0$

$\therefore x-3y+5=0$