Geometry: Circles
Distance to a circle
We have already seen how to calculate the distance between two points and the distance between a point and a line. Now we will see how we can calculate the distance between a point and a circle, two circles, and a line and a circle.
Distance between a point and a circle
Let #\blue c# a circle with centre #M# and radius #\orange r# and #\green P# a point, we distinguish between three cases:
 #d(\green P, \blue c)=\green PM\orange r# if #\green P# lies outside circle #\blue c#.
 #d(\green P, \blue c)=0# if #\green P# lies on circle #\blue c#.
 #d(\green P, \blue c)=\orange r\green PM# if #\green P# lies inside circle #\blue c#.
In the figure drag the point #\green P# to see what effect it has when #\green P# is located inside or outside the circle.
The center of circle #\blue c# can be dragged, and the radius of the circle adjusted with the slider.
The distance between two circles can be calculated by using the centers and their radius.
Distance between two circles
Let #\blue c# be a circle with centre #M# and radius #r_{\blue c}#.
Let #\green d# be a circle with centre #N# and radius #r_{\green{d}}#.
The distance between #\blue c# and #\green d# is equal to:
 #d(\blue c, \green d)=0# if circles #\blue c# and #\green d# touch or intersect.
 #d(\blue c, \green d)=MNr_{\blue c}r_{\green d}# if circles #\blue c# and #\green d# both lie on the outside of the other.
 #d(\blue c, \green d)=r_{\blue c}MNr_{\green d}# if circle #\green d# lies inside circle #\blue c#.
In the latter situation we can interchange the role of #\blue c# and #\green d# if #\blue c# lies within #\green d#.
Drag the centers #M# and #N# and change the radius with the sliders to see the different situations.
Finally, we can calculate the distance between a line and a circle.
Distance between a line and a circle
Stepbystep 
Example 

We calculate the distance between a line #\green l# and a circle #\blue c#. 
#\blue c: \blue{(x4)^2+(y+3)^2=25}# #\green l: \green{y=x+4}# 

Step 1 
Determine the centre #M# and radius #\orange r# of circle #\blue c#. 
#M=\rv{4,3}# #\orange r=5# 
Step 2 
Determine the equation of the line #m# through the centre #M# and perpendicular to line #\green l#. 
#m: y=x+1# 
Step 3 
Calculate the coordinates of the point intersection #P# of #\green l# and #m#. 
#P=\rv{\frac{3}{2},\frac{5}{2}}# 
Step 4 
We distinguish two cases:

#PM=\frac{11}{2}\sqrt{2} \gt 5# #d(\green l, \blue c)=\frac{11}{2}\sqrt{2}5# 
Circle #c# has center #M=\rv{9, 8}# and radius #r=6#. Therefore, point #P# is outside the circle.
This means #d(P,c)=d(P,M)r#.
We therefore calculate #d(P,M)#. This goes as follows:
\[\begin{array}{rcl}d(P,M)&=&\sqrt{(x_Px_M)^2+(y_Py_M)^2} \\&&\phantom{xxx}\blue{\text{formula distance between two points}} \\
&=& \sqrt{(1+9)^2+(0+8)^2} \\&&\phantom{xxx}\blue{\text{substituted}}\\
&=& \sqrt{164} \\&&\phantom{xxx}\blue{\text{calculated}} \\
\end{array}\]
The radius #r=6# and #d(P,M)=2\cdot \sqrt{41}#, therefore #d(P,c)=2\cdot \sqrt{41}6#.
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