Intersections of a line and a circle

Geometry: Circles

Intersections of a line and a circle

A line and a circle can have two, one, or no points of intersection. In the case that they have one the line is called a tangent line to the circle.

In the figure we see a line and a circle. Move the circle by dragging the center and change the radius by adjusting the slider. Move the line by dragging the two points. See what happens to the number of intersections.

The block describes how to find the intersection of a line and a circle.

Intersections line and circle

	Step-by-step	Example
	We determine the intersections of a line #\blue l# and a circle #\green c#.	#\blue l: \blue{y-2x=4}# #\green c: \green{(x+2)^2+(y-1)^2=1}#
Step 1	Write the equation of line #\blue l# in the form #y=\ldots#	#\blue l: \blue{y=2x+4}#
Step 2	Substitute the equation of line #\blue l#, as found in step 1, in the equation of circle #\green c#.	#(x+2)^2+(2x+4-1)^2=1#
Step 3	expand the brackets of quadratic equation in #x# from step 2 and determine the discriminant #D \lt 0# then #\blue l# and #\green c# have no point of intersection and we're done. #D=0# then #\blue l# and #\green c# have one point of intersection, proceed to step 4. #D \gt 0# then #\blue l# and #\green c# have two points of intersection, proceed to step 4.	#5x^2+16x+12=0# # \begin{array}{rcl}D&=&16^2-4 \cdot 5 \cdot 12\\&=&16 \\ &\gt& 0\end {array} # We find two points of intersection.
step 4	Determine the #x# coordinates of the intersection points by solving the quadratic equation in step 2 using the quadratic formula.	#x=\tfrac{-16-\sqrt{16}}{2 \cdot 5} \lor \tfrac{-16+\sqrt{16}}{2\cdot 5}# # x=-2 \lor x=-\tfrac{6}{5}#
step 5	Substitute the #x# coordinates found in step 4 in the equation of the line from step 1 to determine the corresponding #y# coordinates.	If #x=-2#, then #y=2 \cdot -2+4=0# If #x=-\tfrac{6}{5}#, then #y=2 \cdot -\tfrac{6}{5}+4=\frac{8}{5}#

Picture

\[\phantom{x}\]

In the picture, line #\blue l: \blue{y-2x=4}# and circle #\green c: \green{(x+2)^2+(y-1)^2=1}# are drawn.

Interchange role x and y

In step #1# we can also rewrite #l# to the form #x=\ldots#. The rest of the steps will then be the same with the role of #x# and #y# interchanged. This means that every #x# should be replaced with a #y# and vice versa. This way we can also find the points of intersection with a vertical line.

Determine the points of intersection of the line #l: y=7\cdot x+3# with the circle #c: \left(x-1\right)^2+\left(y-3\right)^2=81#.

Give your answer in the form

#none# #\phantom{xxxwwxx}# if there is no point of intersection,
#\left\{\rv{a,b}\right\}\phantom{xxxww}# if there is one point of intersection and
#\left\{\rv{a,b},\rv{c,d}\right\}\phantom{x}# if there are two points of intersection ,

in which #a#, #b#, #c#, #d# are exact numbers.

#\left\{\rv{{{1-\sqrt{4001}}\over{50}},3-{{7\cdot \left(\sqrt{4001}-1\right)}\over{50}}},\rv{{{\sqrt{4001}+1}\over{50}},{{7\cdot \left(\sqrt{4001}+1\right)}\over{50}}+3}\right\}#

Step 1	The line #l# is already in the form #y=\ldots#.
Step 2	We substitute the equation of line #l# in the equation of the circle. That gives: \[\left(x-1\right)^2+\left(7\cdot x+3-3\right)^2=81\] This can be simplified to: \[49\cdot x^2+\left(x-1\right)^2=81\]
Step 3	We reduce the equation from step 2 to #0# and expand the brackets. This goes as follows: \[\begin{array}{rcl}49\cdot x^2+\left(x-1\right)^2&=&81 \\&&\phantom{xxx}\blue{\text{original equation}} \\ 50\cdot x^2-2\cdot x+1&=&81\\&&\phantom{xxx}\blue{\text{expanded brackets}} \\ 50\cdot x^2-2\cdot x-80 &=& 0 \\&&\phantom{xxx}\blue{\text{reduced to }0} \ \end{array}\] We now read #a#, #b# and #c# for the quadratic formula. That gives: #a=50#, #b=-2# and #c=-80#. We can now calculate the discriminant. \[\begin{array}{rcl}D&=&b^2-4ac \\&&\phantom{xxx}\blue{\text{formula discriminant}} \\ &=& (-2)^2-4\cdot 50 \cdot -80 \\&&\phantom{xxx}\blue{\text{substituted}} \\ &=& 16004 \end{array}\] Since the discriminant is equal to # 16004 \gt 0 #, there are #2# solutions.
Step 4	We continue to solve the equation using the quadratic formula. \[\begin{array}{rcl}x=\frac{-b-\sqrt{D}}{2a} &\lor& \frac{-b+\sqrt{D}}{2a} \\&&\phantom{xxx}\blue{\text{quadratic formula}} \\ x=\frac{-{-2}-\sqrt{16004}}{2 \cdot 50} &\lor&\frac{-{-2}+\sqrt{16004}}{2 \cdot 50} \\&&\phantom{xxx}\blue{\text{substituted in quadratic formula}} \\ x={{1-\sqrt{4001}}\over{50}} &\lor& x={{\sqrt{4001}+1}\over{50}} \\&&\phantom{xxx}\blue{\text{simplified}} \end{array}\]
Step 5	Now we determine the appropriate #y#-values found by substituting the #x#-values in the equation of line #l#. For #x={{1-\sqrt{4001}}\over{50}}#, it holds that # y = 7 \cdot {{1-\sqrt{4001}}\over{50}} +3 = 3-{{7\cdot \left(\sqrt{4001}-1\right)}\over{50}} #. For #x={{\sqrt{4001}+1}\over{50}}#, it holds that # y = 7 \cdot {{\sqrt{4001}+1}\over{50}} +3 = {{7\cdot \left(\sqrt{4001}+1\right)}\over{50}}+3 #. The points of intersection are therefore: \[\left\{\rv{{{1-\sqrt{4001}}\over{50}},3-{{7\cdot \left(\sqrt{4001}-1\right)}\over{50}}},\rv{{{\sqrt{4001}+1}\over{50}},{{7\cdot \left(\sqrt{4001}+1\right)}\over{50}}+3}\right\}\]

New example