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

Question

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.

Answer 1

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Answer 2

To show that the equation of the circle C can be written as r = 2s*cos(theta + t), we can use the general equation of a circle centered at (h, k) with radius R:

(x - h)^2 + (y - k)^2 = R^2.

In this case, since the circle passes through the origin (0, 0), we have:

x^2 + y^2 = R^2.

Now, let's convert this equation into polar coordinate form. In polar coordinates, x = r*cos(theta) and y = r*sin(theta). Substituting these values into the equation, we get:

(r*cos(theta))^2 + (r*sin(theta))^2 = R^2.

Applying the trigonometric identity cos^2(theta) + sin^2(theta) = 1, we simplify the equation to:

r^2(cos^2(theta) + sin^2(theta)) = R^2.

Since cos^2(theta) + sin^2(theta) = 1, we have:

r^2 = R^2.

Now, notice that r = 2s*cos(theta + t) can be rewritten as r^2 = (2s*cos(theta + t))^2.

Expanding the square, we have:

r^2 = 4s^2*cos^2(theta + t).

Comparing this equation with r^2 = R^2, we find that R^2 = 4s^2 and cos^2(theta + t) = 1.

Therefore, we have R = 2s and cos(theta + t) = 1.

Since cos(theta + t) equals 1 when theta + t is an integer multiple of 2π, we can write:

theta + t = 2nπ, where n is an integer.

Solving for t, we find:

t = -theta + 2nπ.

So, we can conclude that the equation of the circle C is r = 2s*cos(theta + t), where s = R/2 and t = -theta + 2nπ (n is an integer).

Answer 3

To show that the equation r = 2s*cos(theta + t) represents a circle passing through the origin, we can rewrite it in the form (x-h)^2 + (y-k)^2 = R^2, where (h, k) represents the center of the circle and R represents the radius.

Given the polar equation r = 2s*cos(theta + t), we can convert it to Cartesian coordinates using the relationships x = r*cos(theta) and y = r*sin(theta). Substituting these values, we get:

x = 2s*cos(theta + t)*cos(theta)
y = 2s*cos(theta + t)*sin(theta)

Now, we can square both equations:

x^2 = 4s^2*cos^2(theta + t)*cos^2(theta)
y^2 = 4s^2*cos^2(theta + t)*sin^2(theta)

Since cos^2(theta) + sin^2(theta) = 1, we can rewrite the above equations as:

x^2 = 4s^2*cos^2(theta + t)*(1 - sin^2(theta))
y^2 = 4s^2*cos^2(theta + t)*sin^2(theta)

Using the trigonometric identity cos^2(theta) = 1 - sin^2(theta), we can simplify further:

x^2 = 4s^2*cos^2(theta + t)*(1 - sin^2(theta))
= 4s^2*(1 - sin^2(theta + t))*(1 - sin^2(theta))
= 4s^2*(1 - sin^2(theta + t) - sin^2(theta) + sin^2(theta + t)*sin^2(theta))
= 4s^2*(1 - sin^2(theta + t) - sin^2(theta) + sin^2(theta)*cos^2(theta + t))

By using the double-angle identity sin(2theta) = 2sin(theta)*cos(theta), we can simplify further:

x^2 = 4s^2*(1 - sin^2(theta + t) - sin^2(theta) + (sin^2(theta))*(1 - sin^2(theta + t)))
= 4s^2*(1 - sin^2(theta + t) - sin^2(theta) + sin^2(theta) - sin^4(theta + t))
= 4s^2*(1 - sin^4(theta + t) - sin^2(theta) + sin^2(theta) - sin^4(theta + t))
= 4s^2*(1 - 2sin^4(theta + t) - sin^2(theta))

Similarly, we can simplify the equation for y^2:

y^2 = 4s^2*cos^2(theta + t)*sin^2(theta)
= 4s^2*(1 - sin^2(theta + t))*sin^2(theta)
= 4s^2*(sin^2(theta) - sin^4(theta + t))

Combining the x and y equations, we get:

x^2 + y^2 = 4s^2*(1 - 2sin^4(theta + t) - sin^2(theta) + sin^2(theta) - sin^4(theta + t) + sin^2(theta) - sin^4(theta + t))
= 4s^2*(1 - 4sin^4(theta + t))
= R^2

Thus, we have shown that the equation r = 2s*cos(theta + t) represents a circle passing through the origin, with radius R = 2s.

In conclusion, to find real numbers s and t such that C is the graph of r = 2s*cos(theta + t), you can use the method explained above to derive the equation of a circle in Cartesian coordinates. By equating the resulting equation to (x-h)^2 + (y-k)^2 = R^2, you can determine the values of h, k, and R, which will correspond to the center and radius of the circle.

Answer 4

consider the circle

r = 2cosθ
that is (x-1)^2+y^2 = 1

clearly
r = 2scosθ

is just a scaled up version of that.

(x - s)^2 + y^2 = s^2

r = 2scos(θ+t)

is just a rotated version, turned through angle t.

Since t can be anything from 0 to 2pi, the center of our circle can be anywhere on the circle (x-s)^2+y^2 = s^2

or something.