solve the given quadratic equation for x:
. Use the formula to solve a quadratic equation:
. There exists only one solution if the radicand equals zero:
From this equation follows:
If the radicand equals zero, than the x-value of the tangent point is
. Now use the term to substitute (1+mē) and you'll get:
. Plug in this term into the equation of the tangent:
From the equation of part (a) follows:
Use this result to simplify the y-value to:
The slope of the line from the centre to the tangent point is:
Two lines are perpendicula if their slopes hold the equation:
. Thus the tangent and the line containing centre and tangent point are perpendicular.
There are 2 different situations:
(i) You already know the tangent point. Then draw the line centre-tangentpoint and afterwards a perpendicular line through the tangent point.
(ii) The tangent passes through a point outside the circle. Then you have to construct a right triangle using the circle of Thales.