Question 1207940

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
{{{x^2+y^2=r^2}}}
The equation of the tangent line is given
y=mx+b
substitute y= mx+b in the equation of circle
{{{ x^2 + (mx+b)^2 = r^2}}}
{{{ x^2 + m^2x^2+2mbx +b^2 = r^2}}}
{{{(m^2+1)x^2+ 2mbx +(b^2-r^2)=0}}}
This is a quadratic equation and since tangent and circle meet at only one point there is only one solution
Discriminant {{{b^2-4ac =0}}}
Substitute b ,a and c  from the above equation .
{{{(2mb)^2  -  4((m^2+1)(b^2-r^2))=0}}} 
{{{ 4m^2b^2-4(m^2b^2-m^2r^2+b^2-r^2)=0}}}
{{{ 4m^2b^2-4m^2b^2+4m^2r^2-4b^2+4r^2)=0}}}
{{{ 4m^2r^2-4b^2+4r^2=0}}}
{{{ 4m^2r^2+4r^2=4b^2}}}
{{{r^2(m^2+1) = b^2 }}}-----------------------------------------A
B
{{{(m^2+1)x^2+ 2mbx +(b^2-r^2)=0}}}
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

{{{r^2(m^2+1) = b^2 }}    from A
{{{(m^2+1)= b^2/r^2}}}
Substitute(m^2+1)  in 1
y = b/(b^2/r^2)
y=r^2/b
{{{x = -mb/(m^2+1) = -mb/(b^2/r^2) = -(mr^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