Question... Jesus, Mary and Joseph! (1 Viewer)

[currysauce]

Sweet Merciful Crap... try this

The line l : y=mx-mb through the point P(b,0) outside the circle x²+y²=1 meets the circle at the points A and B with x-coordinates (alpha and beta).

(a)
Show that alpha and beta satisfy the equation (m²+1)x² - 2m²bx + (m²b²-1) = 0

(b)
Show that if l is a tangent to the circle, then m²(b²-1) = 1. Hence find the eqation of the line ST joining the points S and T of the tangents to the Circle from P.

(c)
The general line l meets ST at Q. Prove that Q divides AB internally in the same ratio as P divides AB externally.

 

[Slidey]

The line l : y=mx-mb through the point P(b,0) outside the circle x²+y²=1 meets the circle at the points A and B with x-coordinates (alpha and beta).

mx-mb=sqrt(1-x^2)
m^2(x^2-2bx+b^2)=1-x^2
(m^2+1)x^2-2bm^2x+b^2m^2-1=0
Since alpha and beta are the points of intersection of both functions, and both functions solve simulataneously to produce the above equation, alpha and beta must be roots of the above equation.

"(b)
Show that if l is a tangent to the circle, then m²(b²-1) = 1."

(m²+1)x² - 2m²bx + (m²b²-1) = 0
If L is a tangent to the circle, than alpha and beta are equal, hence the discriminant must equal zero.
D=b²-4ac=0
D=4m^4b^2-4(m^2+1)(m^2b^2-1)=0
D=m^4b^2-(m^4b^2+m^2b^2-1-m^2)=0
D=m^2b^2-1-m^2=0
m^2(b^2-1)=1 #

"Hence find the eqation of the line ST joining the points S and T of the tangents to the Circle from P."

Hmm. Maybe you need to look at something about circle geometry regarding tangents and chords.

I'll have a further look in the morning.

 

[illin]

Ive done this question before, if memory serves me correctly...

a) substitute y=m(x-b) into the circle equation
lots of algrebra needed, the only way to do it

b) discriminant=0, only one solution to part a) line is a tangent
solve for this

c) is easy
use circle geometry