1. ## Quadratic Polynomials

Hi

Could someone please help me get started on this question. I seriously have no clue how to do part (i).

Given that y=mx + b is a tangent to the circle x^2 + y^2 = a^2.

(i) Show that b^2 = a^2 (m^2 +1)

(ii) If y = 3x + b and x^2 + y^2 = 25, find b.

2. On the circle
$\displaystyle x^2+y^2=a^2$

differentiating:

$\displaystyle 2xdx+2ydy=0$

So

dy/dx=-x/y on the circle (slope of tangent at arbitary point).

So if we call the point of intersection between the circle and the line $\displaystyle x_0,y_0$ then the tangent line has slope $\displaystyle m=-x_0/y_0$ and passes through the point $\displaystyle x_0,y_0$.

That is:
$\displaystyle y_0=mx_0+b$

$\displaystyle b=y_0-mx_0$

$\displaystyle b^2=y_0^2+m^2x_0^2-2mx_0$

$\displaystyle b^2=y_0^2+m^2x_0^2-2m(x_0/y_0)y_0^2$

$\displaystyle b^2=y_0^2+m^2x_0^2+2m^2y_0^2$

$\displaystyle b^2=y_0^2+m^2y_0^2+m^2(x_0^2+y_0^2)$

$\displaystyle b^2=y_0^2+(x_0^2/y_0^2)y_0^2+m^2(x_0^2+y_0^2)$

$\displaystyle b^2=y_0^2+x_0^2+m^2(x_0^2+y_0^2)$

$\displaystyle b^2=a^2+m^2a^2$

$\displaystyle b^2=a^2(1+m^2)$ as required