Answer
Tangent touches circle at one point only, using this condition we can prove the condition $r^2(1+m^2) = b^2$
Point of contact = $(\frac{-r^2m}{b}, \frac{r^2}{b})$
Radius to the point of contact and tangent is perpendicular to each other
Work Step by Step
Given equation of
Circle $x^2 +y^2 = r^2$
Tangent $y = mx +b$
Condition that should be met is $r^2(1+m^2) = b^2$
Lets solve for point of contact between tangent and circle
Putting $y = mx +b$ in the equation of circle we get
$x^2 + (mx+b)^2 = r^2$
$=> (1+m^2)x^2 +2mbx + b^2 -r^2 = 0$ - - - -- (i)
Now tangent touches circle at just one point, so we should get only one solution from the quadratic
Therefore discriminant $b^2 -4ac = 0$
=> $(2mb)^2 - 4(1+m^2)(b^2-r^2)=0$
=> $4m^2b^2 - 4b^2 + 4r^2 - 4m^2b^2+4m^2r^2 = 0$
=>$ 4r^2 +4m^2r^2 = 4b^2$
=>$r^2(1+m^2) = b^2$ proved
b) Point of tangency
x will be equal solution of Quadratic equation (i)
$x = \frac{-2mb}{2(1+m^2)}$
Using $r^2(1+m^2) = b^2$
we get $x = \frac{-r^2m}{b}$
Putting x in the tangent equation $y = mx +b$
we get $y = \frac{-r^2m}{b}m +b = \frac{-r^2m^2}{b} +b$
Using $r^2(1+m^2) = b^2$
we get $y = \frac{r^2}{b}$
Point of touching $(\frac{-r^2m}{b}, \frac{r^2}{b})$
c) To prove tangent is perpendicular to line joining centre and point of contact
slope of tangent $m_1 = m$
Slop of radius $m_2 = \frac{\frac{r^2}{b}}{\frac{-r^2m}{b}} = \frac{-1}{m}$
we see $m_1m_2 = -1$, so they are perpendicular
