Tangent line to a circle

Geometry: Circles

Tangent line to a circle

When a line and a circle have exactly one point of intersection, the line is called a tangent to the circle.

Tangent to a circle

If a line #\blue l# and circle #\green c# have exactly one point of intersection #\orange T# then #\blue l# is called the tangent to #\green c# at #\orange T#.

We call the point #\orange T# the tangent point.

In the figure the point #\orange T# can be moved across the circle #\green c#. Circle #\green c# can be moved by dragging its center and the radius can be adjusted using the slider.

Each tangent has the following property.

Tangent Theorem

A tangent line to a circle is perpendicular to the line through its tangent point and the center of the circle.

We prove the theorem for a circle #\green c# with center #\rv{0,0}# and radius #r# and a tangent line #\blue l# through point #\orange{T}=\orange{\rv{t,s}}# on the circle. Note that moving the center has no effect, so that the statement then remains valid.

The circle equation of #\green c# is equal to #x^2+y^2=r^2#. By means of conversion, we can bring back the circle to two graphs of a function, namely #f(x)=\sqrt{r^2-x^2}# for the positive half of the circle, and #g(x)=-\sqrt{r^2-x^2}# for the negative half of the circle.

Now we can differentiate by calculating the slope of the circle in the point #\orange{T}=\orange{\rv{t,s}}#. The derivative of #f(x)# is equal to #f'(x)=\frac{-2x}{2\sqrt{r^2-x^2}}=\frac{-x}{\sqrt{r^2-x^2}}#. This means that the slope #\orange{T}=\orange{\rv{t,s}}# equals:

\[f'(t)=\frac{-t}{\sqrt{r^2-t^2}}\]

Because the point #\orange{T}=\orange{\rv{t,s}}# lies on the circle, applies \[t^2+s^2=r^2\] This means that #r^2-t^2=s^2#. therefore applies:

\[f'(t)=\frac{-t}{\sqrt{s^2}}=-\frac{t}{s}\]

Similarly we can do this for #g(x)#. Here, paying attention to the equally well to the minus signs, but because #s# is negative, this is the same.

This means that the slope of line #\blue l# also must be equal to #-\frac{t}{s}#. Otherwise #\blue l# not tangent.

Now we calculate the slope of the line #O\orange T#, or the radius of the circle through the point #\orange{T}=\orange{\rv{t,s}}#.

The slope is equal to: \[a_{O\orange T}=\frac{y_T-y_O}{x_T-x_O}=\frac{s-0}{t-0}=\frac{s}{t}\]

We know that two lines perpendicular to each other as their directional multiplied with each coming out on #-1#.

In this case applies to the directional line of #\blue l# and line #O\orange T# :

\[a_{\blue l} \cdot a_{O\orange T}=-\frac{t}{s} \cdot \frac{s}{t}=-1\]

This has proven to be tangent #l# to circle #\green c# in point #\orange{T}=\orange{\rv{t,s}}# perpendicular to the radius to the point of tangency.

We can use the tangent theorem to determine an equation of a tangent to a circle through a given point.

Tangent to a circle

	Step-by-step	Example
	We determine an equation for a tangent line #\blue l# to a circle #\green c# at a point #\orange T#.	#\orange T=\orange{\rv{5,7}}# #\green c : \green{(x-2)^2+(y-3)^2=25}#
Step 1	Determine the centre #M# of circle #\green c#.	#M=\rv{2,3}#
Step 2	Determine the slope of the line through #M# and #\orange T#.	#{a}_{M\orange T}=\tfrac{7-3}{5-2}=\tfrac{4}{3}#
Step 3	Use the tangent theorem to calculate the slope #a# of tangent line #\blue l#.	#a=-\tfrac{3}{4}#
step 4	The equation of line #\blue l# is of the form #y=ax+b#. Substitute the #a# found in step 3.	#\blue l: y=-\tfrac{3}{4}x+b#
step 5	Determine #b# by substituting the point #\orange T# in the equation found in step 4 and solve the resulting equation.	#b=\tfrac{43}{4}#
step 6	Substitute #b# into the equation of step 4. This gives an equation for line #\blue l#.	#\blue l: y=-\tfrac{3}{4}x+\tfrac{43}{4}#

Picture

In the figure we see circle \[\green c : \green{(x-2)^2+(y-3)^2=25}\] and point \[\orange T=\orange{\rv{5,7}}\] with the tangent line \[\blue l: y=-\tfrac{3}{4}x+\tfrac{43}{4}\] from the example.

Is line #l: -{{5\cdot x}\over{2}}-5\cdot y={{5}\over{2}}# a tangent to circle #c: \left(x-2\right)^2+\left(y+4\right)^2=5#?

Yes

Line #l# is a tangent to circle #c# when #l# and circle #c# have exactly one intersection point. We therefore determine the number of intersections of line #l# and circle #c#.

Step 1	We rewrite line #l# to form #y=\ldots#. That gives: \[l: y=-{{x}\over{2}}-{{1}\over{2}}\]
Step 2	We substitute the equation of line #l# in the equation of the circle. That gives: \[\left(x-2\right)^2+\left(-{{x}\over{2}}-{{1}\over{2}}+4\right)^2=5\] This can be simplified to: \[\left(x-2\right)^2+\left({{7}\over{2}}-{{x}\over{2}}\right)^2=5\]
Step 3	We reduce the equation from step 2 to #0# and expand the brackets. This goes as follows: \[\begin{array}{rcl}\left(x-2\right)^2+\left({{7}\over{2}}-{{x}\over{2}}\right)^2&=&5 \\&&\phantom{xxx}\blue{\text{original equation}} \\ {{5\cdot x^2}\over{4}}-{{15\cdot x}\over{2}}+{{65}\over{4}}&=&5\\&&\phantom{xxx}\blue{\text{expanded brackets}} \\ {{5\cdot x^2}\over{4}}-{{15\cdot x}\over{2}}+{{45}\over{4}} &=& 0 \\&&\phantom{xxx}\blue{\text{reduced to }0} \ \end{array}\] We now read #a#, #b# and #c# for the quadratic formula. That gives: #a={{5}\over{4}}#, #b=-{{15}\over{2}}# and #c={{45}\over{4}}#. We can now calculate the discriminant. \[\begin{array}{rcl}D&=&b^2-4ac \\&&\phantom{xxx}\blue{\text{formula discriminant}} \\ &=& (-{{15}\over{2}})^2-4\cdot {{5}\over{4}} \cdot {{45}\over{4}} \\&&\phantom{xxx}\blue{\text{substituted}} \\ &=& 0 \end{array}\] Since the discriminant is equal to # 0 #, there is there #1# solution .

Because there is #1# intersection point, line #l# is a tangent to #c#.

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