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# in #\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 can use the tangent theorem to determine an equation of a tangent to a circle through a given point.
Tangent to a circle
Stepbystep 
Example 

We determine an equation for a tangent line #\blue l# to a circle #\green c# and through a point #\orange T#. 
#\orange T=\orange{\rv{5,7}}# #\green c : \green{(x2)^2+(y3)^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#. 
#{rc}_{M\orange T}=\tfrac{73}{52}=\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}# 
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: \[l: y={{38}\over{3}}{{7\cdot x}\over{3}}\] 
Step 2  We substitute the equation of line #l# in the equation of the circle. That gives: \[\left(x3\right)^2+\left({{38}\over{3}}{{7\cdot x}\over{3}}2\right)^2=40\] This can be simplified to: \[\left(x3\right)^2+\left({{32}\over{3}}{{7\cdot x}\over{3}}\right)^2=40\] 
Step 3  We reduce the equation from step 2 to #0# and expand the brackets. This goes as follows: \[\begin{array}{rcl}\left(x3\right)^2+\left({{32}\over{3}}{{7\cdot x}\over{3}}\right)^2&=&40 \\&&\phantom{xxx}\blue{\text{original equation}} \\ {{58\cdot x^2}\over{9}}{{502\cdot x}\over{9}}+{{1105}\over{9}}&=&40\\&&\phantom{xxx}\blue{\text{expanded brackets}} \\ {{58\cdot x^2}\over{9}}{{502\cdot x}\over{9}}+{{745}\over{9}} &=& 0 \\&&\phantom{xxx}\blue{\text{reduced to }0} \ \end{array}\] We now read #a#, #b# and #c# for the quadratic formula. That gives: #a={{58}\over{9}}#, #b={{502}\over{9}}# and #c={{745}\over{9}}#. We can now calculate the discriminant. \[\begin{array}{rcl}D&=&b^24ac \\&&\phantom{xxx}\blue{\text{formula discriminant}} \\ &=& ({{502}\over{9}})^24\cdot {{58}\over{9}} \cdot {{745}\over{9}} \\&&\phantom{xxx}\blue{\text{substituted}} \\ &=& {{2932}\over{3}} \end{array}\] Since the discriminant is equal to # {{2932}\over{3}} \gt 0 #, there are there #2# solutions . 
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