Let C be a circle that passes through the origin. Show that we can find real numbers s and t such that C is the graph of
r = 2s*cos(theta + t)
I think that I could start by writing it like R^2=(x-h)^2+(y-k)^2 where R is the radius of the circle, but I don't what to do with that info.
no help from my earlier comments?
http://www.jiskha.com/display.cgi?id=1461018078
Sorry, they just seem unclear, and you seem unsure of your final answer... And I didn't think you would respond to the original thread so I resposted
Never mind, I think I get it now
good. I thought it would make sense if you looked at it a while.
To show that the equation r = 2s*cos(theta + t) represents a circle passing through the origin, we can rewrite it in terms of Cartesian coordinates (x, y).
Given r = 2s*cos(theta + t), we can convert it to Cartesian coordinates using the trigonometric identities:
x = r*cos(theta) and
y = r*sin(theta).
Substituting the value of r from the given equation, we have:
x = 2s*cos(theta + t)*cos(theta)
y = 2s*cos(theta + t)*sin(theta)
Simplifying these equations, we get:
x = 2s*cos(theta)*cos(t) - 2s*sin(theta)*sin(t)
y = 2s*cos(theta)*sin(t) + 2s*sin(theta)*cos(t)
Now, we can apply the Pythagorean identity cos^2(x) + sin^2(x) = 1, which implies:
cos^2(t) + sin^2(t) = 1
cos^2(t) = 1 - sin^2(t)
Let's substitute these identities into the equations for x and y:
x = 2s*cos(theta)*(1 - sin^2(t)) - 2s*sin(theta)*sin(t)
y = 2s*cos(theta)*sin(t) + 2s*sin(theta)*cos(t)
Expanding these equations, we get:
x = 2s*cos(theta) - 2s*sin^2(t)*cos(theta) - 2s*sin(theta)*sin(t)
y = 2s*sin(t)*cos(theta) + 2s*cos(t)*sin(theta)
Rearranging the terms in these equations, we have:
x = 2s*cos(theta) - 2s*sin(theta)*sin(t)*cos(theta) - 2s*sin^2(t)*cos(theta)
y = 2s*sin(t)*cos(theta) + 2s*sin(theta)*cos(t)
Now, we can group the terms involving cos(theta) and sin(theta):
x = 2s*(cos(theta) - sin(theta)*sin(t)*cos(theta) - sin^2(t)*cos(theta))
y = 2s*(sin(t)*cos(theta) + sin(theta)*cos(t))
Let A = cos(theta) - sin(theta)*sin(t)*cos(theta) - sin^2(t)*cos(theta)
Let B = sin(t)*cos(theta) + sin(theta)*cos(t)
The equations can then be expressed as:
x = 2s*A
y = 2s*B
These equations show that the circle defined by the equation r = 2s*cos(theta + t) can be written in the Cartesian form x = 2s*A and y = 2s*B, which suggests that it is a circle centered at the origin with radius 2s. Therefore, we have shown that the given equation represents a circle passing through the origin.