helloo i need to get an answer for this question
Show that the line y = mx + c touches the circle x^2 + y^2 = a^2 if c^2 = a^2(1+m^2)
it's very hard for me i need some help plz.
$\displaystyle x^2 + y^2 = a^2$Show that the line y = mx + c touches the circle x^2 + y^2 = a^2 if c^2 = a^2(1+m^2)
$\displaystyle x^2 + (mx + c)^2 = a^2$
$\displaystyle x^2 + m^2 x^2 + c^2+2mcx = a^2$
$\displaystyle (1 + m^2) x^2 +2mcx + (c^2- a^2)=0$
if the line y = mx + c touches the circle(or tangent to the circle) then
$\displaystyle discriminant = 0 $
$\displaystyle d= (2mc)^2 -4(1 + m^2)(c^2- a^2)=0$
$\displaystyle 4a^2-4c^2+4m^2a^2=0$
$\displaystyle c^2 = a^2+a^2m^2$
$\displaystyle c^2 = a^2(1+m^2)$
the equation of the tangent to the circle having centre at origin is given by
$\displaystyle x_{0}x+y_{0}y=r^2$
$\displaystyle where (x_{0},y_{0}) are\ point\ of\ intersection$
$\displaystyle since\ (p,q)\ lies\ on\ the\ circle\ x^2 + y^2 = a^2.$
$\displaystyle therefore\ p^2 + q^2 = a^2 =(radius)^2$
therefore equation of the tangent
$\displaystyle px+qy= p^2 + q^2 $