a straight line passing through the origin with slope m cuts the circle x^2+y^2-4x-2y-4=0 at A(a,b) and B(b,c)

a straight line passing through the origin with slope m cuts the circle x^2+y^2-4x-2y-4=0 at A(a,b) and B(b,c).

A. Find a quadratic equation whose roots are a and c

ans: (1+m^2)x^2-(2m+4)x-4=0

B. if P is the mid-point of AB, find the coordinates of P in terms of m.

ans: P= (m+2)/(1+m^2), (m(m+2))/(1+m^2)

C. as m varies, the locus of P is part of a curve C. Find the equation of C.

ans: x^2+y^2=2x+y

I know how to do question A and question B, but how to do C? any idea?

Top 1 Answers

A few days ago

Anonymous

Favorite Answer

Well the answer is two fold

1. C has a for its divisor the equation

m+0(m*1*0)/m+2*-0 = 3

That is a partial answer to B.

If x=m+3 and y=m than C is the locus of

P-C+m.

I hopes that will help you.

Gloknick Yagoslovisky

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