## Distance Between Points of Intersection of Circle With Axes

Suppose we have a circle that intersects both
$x$
and
$y$
axes. We can find the distanceof the points of intersection of points on the
$x$
axis and points on the
$y$
axis by first finding the points of intersection, then using the distance formula for the distance between two points.
Suppose we have the circle with equation given by
$(x-2)^2+(y-4)^2=25$
. This circle has centre
$(2,4)$ The circle intersects the
$x$
axis when
$y=0$
.
$(x-2)^2+(0-4)^2=25 \rightarrow x-2 =\pm \sqrt{25-16}= \pm 3 \rightarrow x=2-3=-1,2+3=5$
.
The points of intersection with the
$x$
axis are
$(-1,0),(5,0)$
.
The circle intersects the
$y$
axis when
$x=0$
.
$(x-2)^2+(y-4)^2=25 \rightarrow y-4 =\pm \sqrt{25-4}= \pm \sqrt{21} \rightarrow y=4-\sqrt{21}, \: 4+\sqrt{21}$
.
The points of intersection with the
$y$
axis are
$(0, 4-\sqrt{21}),(0,4+\sqrt{21})$
.
The distance between the points closest to the origin,
$(-1,0)$
and
$(4-\sqrt{21},0)$
is
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-1-0)^2+(0-(4-\sqrt{21})^2}=1.157$
to three decimal places. 