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

Corporations compared to partnerships and sole proprietorships?

where are the Aleutian Islands?

EASY But HARD Physics question….?

general from equation?

whats the area of a circle with a radius of 5 inches?

shorthand notes???