Question

Find the equations of circle which passes through the point ( 1 , 1) , (2 , 2) and whose radius is 1 and show that there are two such circles
$(x-2{)}^2+(y-1{)}^2=1$
and $(x-2{)}^2+(y-1{)}^2=1$

Answer 1

Let the centre be (h,k)
$(x-h{)}^2+(y-k{)}^2=1$
or $(1-h{)}^2+(1-k{)}^2=1$
$(2-h{)}^2+(2-k{)}^2=1$, subtract
$(3-2h)\cdot 1+(3-2k)\cdot 1=0$
or $h+3=3\therefore k=3-h$
$(1-h{)}^2+(1-3+h{)}^2=1$
or $(1-h{)}^2+(h-2{)}^2=1$
$2h^2-6h+4=0or{h}^2-3h+2=0$
or $h=1,2\therefore k=2,1$
centres are (1,2) or (2,1) and radius = 1
Ans $(x-1{)}^2+(y-2{)}^2=1$
$(x-2{)}^2+(y-1{)}^2=1$