Question

Find the equation of circle passing through the points $A(1,1)$ & $B(2,2)$ and whose radius is $1$

Answer 1

Equation of AB is $x-y=0$ $\therefore$ equation of circle is
$(x-1)(x-2)+(y-1)(y-2)+$ $\lambda$$(x-y)=0$
or $x^2+y^2+(\lambda -3)x-(\lambda +3)y+4=0$
radius = $\sqrt{\frac{(\lambda -3{)}^2}{4}+\frac{(\lambda +3{)}^2}{4}-4}$ But radius = 1 (given)
$\therefore \sqrt{\frac{(\lambda -3{)}^2}{4}+\frac{(\lambda +3{)}^2}{4}-4}=1or(\lambda -3{)}^2+(\lambda +3{)}^2-16=4\Rightarrow \lambda =\pm 1$
$\therefore$ equation of circles are $x^2+y^2-2x-4y+4=0\&x^2+y^2-4x-2y+4=0$