Question

Using distance formula, show that $(3,3)$ is the center of the circle passing through the points $(6,2),(0,4)$ and $(4,6)$.

Answer 1

Let $C(3,3)$ is the centre of the circle passing through the points $P(6,2),Q(0,4)$ and $R(4,6)$.
$CP=CQ=CR$ [radii of same circle]
By distance formula, $CP$ = $\sqrt{}\left[\right(x2-x1{)}^2+(y2-y1{)}^2]$
$CP$ =$\sqrt{}\left[\right(6-3{)}^2+(2-3{)}^2]$
$CP$ = $\sqrt{}\left[\right(3{)}^2+(-1{)}^2]$
$CP$ = $\sqrt{}[9+1]$
$CP$ = $\sqrt{10}$
By distance formula, $CQ$ = $\sqrt{}\left[\right(x2-x1{)}^2+(y2-y1{)}^2]$
$CQ$ = $\sqrt{}\left[\right(0-3{)}^2+(4-3{)}^2]$
$CQ$ = $\sqrt{}\left[\right(3{)}^2+\left(1{)}^2\right]$
$CQ$ = $\sqrt{}[9+1]$
$CQ$ = $\sqrt{}10$
By distance formula, $CR$ =$\sqrt{}\left[\right(x2-x1{)}^2+(y2-y1{)}^2]$
$CR$ = $\sqrt{}\left[\right(4-3{)}^2+(6-3{)}^2]$
$CR$ = $\sqrt{}\left[\right(1{)}^2+\left(3{)}^2\right]$
$CR$ = $\sqrt{}[1+9]$
$CR$ = $\sqrt{}10$
Since $CP=CQ=CR,$
$C(3,3)$ is the centre of the circle passing through the points $P(6,2),Q(0,4)$ and $R(4,6)$.
Hence proved.