Find the equation of the circle which passes through the points A(1, 1) and B(2, 2) and whose radius is 1. Show that there are two such circles.
The general equation of a circle: $(x-h)^{2}+(y-k)^{2}=r^{2}$
$\ldots(\mathrm{i})$, where $(\mathrm{h}, \mathrm{k})$ is the centre and $\mathrm{r}$ is the radius.
Putting $A(1,1)$ in (i)
$(1-h)^{2}+(1-k)^{2}=1^{2}$
$\Rightarrow h^{2}+k^{2}+2-2 h-2 k=1$
$\Rightarrow h^{2}+k^{2}-2 h-2 k=-1 . .$ (ii)
Putting $B(2,2)$ in (i)
$(2-h)^{2}+(2-k)^{2}=1^{2}$
$\Rightarrow h^{2}+k^{2}+8-4 h-4 k=1$
$\Rightarrow h^{2}+k^{2}-4 h-4 k=-7$
$\Rightarrow\left(h^{2}+k^{2}-2 h-2 k\right)-2 h-2 k=-7$
$\Rightarrow-1-2 h-2 k=-7[$ from (ii) $]$
$\Rightarrow-2 h-2 k=-6$
$\Rightarrow h+k=3 \Rightarrow h=3-k$
Putting it in (ii)
$\Rightarrow(3-k)^{2}+k^{2}-2(3-k)-2 k=-1$
$\Rightarrow 2 k^{2}+4-6 k=0$
$\Rightarrow k^{2}-3 k+2=0$
$\Rightarrow \mathrm{k}=2$ or $\mathrm{k}=1$
When $k=2, h=3-2=1$
Equation of 1 circle
$(x-1)^{2}+(y-2)^{2}=1$
When $\mathrm{k}=1, \mathrm{~h}=3-1=2$
$(x-2)^{2}+(y-1)^{2}=1$
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