AT is a tangent at T to a circle, centre O.
OT = x cm AT = (x+5) cm OA = (x+8) cm
a) Show that x^2-6x-39=0
b) Solve the equation x^2-6x-39=0 to find the radius of the circle

2. Originally Posted by alice
AT is a tangent at T to a circle, centre O.
OT = x cm AT = (x+5) cm OA = (x+8) cm
a) Show that x^2-6x-39=0
I think you probably meant to say: Show that $(x+8)^2=x^2+(x+5)^2$

$x^2+16x+64=x^2+x^2+10x-25$

$x^2-6x-39=0$
Originally Posted by alice
b) Solve the equation x^2-6x-39=0 to find the radius of the circle

I got $x=\frac{6+\sqrt{192}}{2}\approx 9.93$