SOLUTION: The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. If the equation of the circle is x^2 + y^2 = r^2

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Question 1207940: The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:

A. r^2(1 + m^2) = b^2

B. The point of tangency is [(-r^2 m)/b, (r^2/b)]

C. The tangent line is perpendicular to the line containing the center of the circle and point of tangency.


Found 2 solutions by Edwin McCravy, mananth:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
We find the x-intercept of y = mx + b, by setting y = 0.
y = mx + b
0 = mx + b
-b = mx

 


All 6 triangles ACO, AOB, OCB, ACO, ADC, DOC are similar, because a perpendicular
drawn from the right angle to the hypotenuse divides a right triangle into
two right triangles, each similar to it.

 








And by the Pythagorean theorem:




So equating expressions for BC2







----------------------

For the coordinates of the point of tangency, C.  
By similar triangles,



  <--the x-coordinate of the point of tangency C



  <--the y-coordinate of the point of tangency C

So the point of tangency C is 

Edwin
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!

Since tangent and circle meet at only one point there is only one solution . Let the point be (x,y)
The equation of the circle is given

The equation of the tangent line is given
y=mx+b
substitute y= mx+b in the equation of circle



This is a quadratic equation and since tangent and circle meet at only one point there is only one solution
Discriminant
Substitute b ,a and c from the above equation .





-----------------------------------------A
B

Since the discriminant is zero, the equation has exactly one solution for x. The solution for x in a quadratic equation is given by
x = -b/2a
Here it is -2mb/2(m^2+1) = -mb/(m^2+1)
substitute this value of x into the equation y=mx+b
y = m(-mb/(m^2+1)) +b )
y = (-m^2b/(m^2+1)) +b )
y= (-m^2b +m^2b+b)/(m^2+1)
y = b/(m^2+1)-----------------------------------------------1

Substitute(m^2+1) in 1
y = b/(b^2/r^2)
y=r^2/b

The point of tangency is (-r^2 m)/b, (r^2/b) ……………………….B
C
We have to prove tangent and radius are perpendicular.
From circle equation we know co ordinates of centre are (0,0)
The point of contact of radius and tangent are (-r^2 m)/b, (r^2/b)
Find slope using two point formula
Slope = ((r^2/b))/((-r^2m/b))
Slope =(- 1/m)
Slope of tangent line = m
M*(-1/m)= -1
Hence The tangent line is perpendicular to the line containing the center of the circle and point of tangency………….C




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