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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  
\[(2,4)\]
  and radius 5.

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.