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?
To find the equation of the curve C, we need to eliminate the parameter m and express x and y solely in terms of each other.
From part B, we have the coordinates of point P as P = [(m+2)/(1+m^2), (m(m+2))/(1+m^2)].
Let's start by eliminating m. We can rewrite the coordinates of point P as:
x = (m+2)/(1+m^2)
y = (m(m+2))/(1+m^2)
Now, we can use these equations to eliminate m. Multiply the equation for x by m^2 and the equation for y by 1 + m^2:
m^2 * x = (m+2) * m^2 / (1+m^2)
(1+m^2) * y = (m(m+2))
Expanding these equations, we get:
m^2 * x = (m^3 + 2m^2) / (1+m^2)
y + m^2 * y = m^2 + 2m
Now, let's isolate the variables x and y:
m^2 * x = (m^3 + 2m^2) / (1+m^2)
x = (m^3 + 2m^2) / (m^2 + 1)
y + m^2 * y = m^2 + 2m
y * (1 + m^2) = m^2 + 2m
y = (m^2 + 2m) / (1 + m^2)
Now, substitute these expressions for x and y back into the equation x^2 + y^2 = 2x + y:
[(m^3 + 2m^2) / (m^2 + 1)]^2 + [(m^2 + 2m) / (1 + m^2)]^2 = 2[(m^3 + 2m^2) / (m^2 + 1)] + [(m^2 + 2m) / (1 + m^2)]
Simplifying this equation, we get:
m^6 + 3m^4 + 3m^2 + 1 + m^4 + 4m^3 + 4m^2 + m^2 + 4m + 4 = 2m^3 + 4m^2 + 2m + m^2 + 2m
Combining like terms, we have:
m^6 + 4m^4 + 6m^3 + 6m^2 + 5m + 5 = 0
Thus, the equation of the curve C is:
m^6 + 4m^4 + 6m^3 + 6m^2 + 5m + 5 = 0