Geometry: Circles
Intersections of circles
We have seen that we can calculate the points of intersection of a line and a circle. Two circles can also have zero, one or two points of intersections.
In the figure we see a circle #\blue c# with centre #M# and a circle #\green d# with centre #N#. Change the radius of the circles with the sliders or drag the center of the circles and see what happens with the number of points of intersections.
The block below shows how to find the intersection points of two circles.
Calculating points of intersection of circles
Stepbystep 
Example  
We calculate the coordinates of the points of intersections of two circles #\blue c# and #\green d#. 
#\blue c : \blue{x^2+y^2=25}# #\green d : \green{(x1)^2+(y4)^2=16}# 

Step 1 
Expand the brackets in both equations. 
#\blue c : \blue{x^2+y^2=25}# #\green d : \green{x^22x+y^28y=1}# 
Step 2 
Eliminate #x^2+y^2# by subtracting the equations of the circles. The resulting equation is the equation for the line through the points of intersection. 
#2x+8y=26# 
Step 3  Calculate the points of intersection of the line in step 2 with one of the circles. 
#\rv{3,4}# #\rv{\tfrac{77}{17}, \tfrac{36}{17}}# 
We can now calculate the intersection points of two circles. We have already seen how we can create the tangent at a point #T# on the circle. Now we will see how we can determine an equation for a tangent line to a circle with a point outside the circle.
Tangent to a circle from a point outside the circle
Stepbystep 
Example 

We determine equations for the tangent lines to a circle #\blue c# through a point #\green P# outside the circle. 
#\blue c : \blue{(x3)^2+(y1)^2=9}# #\green P=\green{\rv{9,1}}# 

Step 1 
Determine the centre #M# of circle #\blue c#. 
#M=\rv{3,1}# 
Step 2 
Determine the length of #M\green P#. 
#d(M, \green P)=6# 
Step 3 
Suppose the tangent points are called #\orange A# and #\orange B#. Determine length #\orange A\green P# (and thus also #\orange B \green P# ) using the Pythagorean theorem in the rightangled triangle #\triangle \green P M\orange A#. 
#d(\orange A, \green P)=\sqrt{27}# 
Step 4 
Circle #d# is the circle with center #\green P# passing through #\orange A# and #\orange B#. Determine an equation of circle #d# with the help of the radius found in step 3. 
#d: (x9)^2+(y1)^2=27# 
Step 5 
Calculate the coordinates of the points #\orange A# and #\orange B#. They are the intersection points of the circles #\blue c# and #d#. 
#\orange A=\rv{\frac{9}{2}, \frac{3\sqrt{3}2}{2}}# #\orange B=\rv{\frac{9}{2}, \frac{3\sqrt{3}2}{2}}# 
Step 6 
Use the coordinates of step 5 and the coordinates of #\green P# to determine equations for the tangent lines #\green P \orange A# and #\green P \orange B#. 
#\green P \orange A: \frac{1}{\sqrt{3}}x+13\sqrt{3}# #\green P \orange B: \frac{3\sqrt{3}4}{2}x+3\sqrt{3}3# 
Give your answer in the form
 #none# #\phantom{xxxxxvw}# if there is no point of intersection,
 #\left\{\rv{a_1,b_1}\right\}\phantom{xxxww}# if there is one point of intersection and
 #\left\{\rv{a_1,b_1},\rv{a_2,b_2}\right\}\phantom{x}# if there are two points of intersection,
Step 1  We first expand the brackets in both circle equations. That gives the following circle equations: \[c: x^2+12\cdot x+y^2+{{297\cdot y}\over{4}}+{{90513}\over{64}}={{ 119969}\over{64}}\] and \[d: x^22\cdot x+y^2+{{73\cdot y}\over{4}}+{{5393}\over{64}}={{13345 }\over{64}}\] 
Step 2  Now we eliminate #x^2+y^2# by calculating #cd#. That gives: \[56\cdot y+14\cdot x+1330=1666\] This can be reduced to the line: \[y=6{{x}\over{4}}\] 
Step 3  We now calculate the points of intersection of the line found in step 2, and one of the circles. We choose circle #c#. Substituting the line #y=6{{x}\over{4}}# in circle #c# gives: \[\left(x+6\right)^2+\left(6{{x}\over{4}}+{{297}\over{8}}\right)^2={{119969}\over{ 64}}\] We expand the brackets in this equation and reduce to #0#. We then get the equation: \[{{17\cdot x^2}\over{16}}{{153\cdot x}\over{16}}+{{85}\over{4}}=0\] This is a quadratic equation that we can solve using the quadratic formula or by factorization. The solutions are: \[x=4 \lor x=5\] To find the corresponding #y#values, we substitute the found #x#values in the equation of the line. This gives #x=4#: \[y=6{{4}\over{4}}=5\] For #x=5# it gives \[y=6{{5}\over{4}}={{19}\over{4}}\] The points of intersection are therefore: \[\left\{\rv{4,5},\rv{5,{{19}\over{4}}}\right\}\] 
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