circles and limits

[pradababe128]

i was given this problem for hw and i just can not figure out where to start!

(i) In a Cartesian coordinate plane, let A : (0, 0), B : (3, 0). Consider the set all
points C in the plane such that AC/BC = 2. Prove that the set is a circle. Find the
coordinates of its center and its radius length.
Hint: Let C : (x, y).
(ii) Let A and B be two distinct points in the plane. For a ?xed real number k,
0 < k (and k does not equal 1), consider the set all points C in the plane such that AC/BC = k.
Prove that the set is a circle. Describe the position of the center and its radius length as functions
of k.
Hint: Introduce a Cartesian coordinate system in the plane such that A becomes the
origin, and B : (0, 1). Let C : (x, y).

(iii) Let (a, b) be the coordinates of the center of the circle from part (ii), when the
coordinate system was chosen as was suggested in the hint, and r be the length of the
radius. Then a = a(k), b = b(k), and r = r(k). What can be said about
lim
k ?1 a(k),
lim
k ?1 b(k), and
lim
k ?1 r(k)?
What can be said about
lim a(k)
k ?0+ ,
lim b(k)
k?0+ and
lim
k?0+ r(k)?

 

[fasteddie65]

Here's how to do part i:

(i) In a Cartesian coordinate plane, let A : (0, 0), B : (3, 0). Consider the set all
points C in the plane such that AC/BC = 2. Prove that the set is a circle. Find the
coordinates of its center and its radius length.
Hint: Let C : (x, y).

AC = ?((x - 0)^2 + (y - 0)^2)) = ?(x^2 + y^2)

BC = ?((x - 3)^2 + (y - 0)^2) = ?((x - 3)^2 + y^2)

AC/BC = ?(x^2 + y^2)/?((x - 3)^2 + y^2) = 2

?(x^2 + y^2) = 2?((x - 3)^2 + y^2)

x^2 + y^2 = 4[(x - 3)^2 + y^2]

x^2 + y^2 = 4(x^2 - 6x + 9) + 4y^2
x^2 + y^2 = 4x^2 - 24x + 36 + 4y^2

0 = 3x^2 - 24x + 36 + 3y^2

x^2 - 8x + 12 + y^2 = 0

x^2 - 8x + 16 + y^2 = 4

(x - 4)^2 + y^2 = 4

This is an equation of a circle with center (4, 0) and radius 2.

 

[BigGlenntheHeavy]

Part 2

\(\displaystyle Given: \ A0,0), \ B0,1), \ Cx,y), \ \frac{AC}{BC} \ = \ K, \ K>0, \ K\ne 1\)

\(\displaystyle Distance \ of \ AC \ = \ [(x-0)^{2}+(y-0)^{2}]^{1/2} \ = \ [x^{2}+y^{2}]^{1/2}\)

\(\displaystyle Distance \ of \ BC \ = \ [(x-0)^{2}+(y-1)^{2}]^{1/2} \ [x^{2}+(y-1)^{2}]^{1/2}\)

\(\displaystyle \frac{AC}{BC} \ = \ K \ = \ \frac{[x^{2}+y^{2}]^{1/2}}{[x^{2}+(y-1)^{2}]^{1/2}}\)

\(\displaystyle Ergo, \ x^{2}+y^{2} \ = \ K^{2}x^{2}+K^{2}y^{2}-2K^{2}y+K^{2}\)

\(\displaystyle x^{2}-K^{2}x^{2}+y^{2}-K^{2}y^{2}+2K^{2}y \ = \ K^{2}\)

\(\displaystyle x^{2}(1-K^{2})+y^{2}(1-K^{2})+2K^{2}y \ = \ K^{2}\)

\(\displaystyle x^{2}+y^{2}+\frac{2K^{2}y}{1-K^{2}} \ = \ \frac{K^{2}}{1-K^{2}}\)

\(\displaystyle x^{2}+\bigg(y+\frac{K^{2}}{1-K^{2}}\bigg)^{2} \ = \ \frac{K^{2}}{1-K^{2}}+\frac{K^{4}}{(1-K^{2})^{2}} \ = \ \frac{K^{2}}{(1-K^{2})^{2}}\)

\(\displaystyle Hence, \ circle \ is \ centered \ at \ \ \bigg(0,-\frac{K^{2}}{1-K^{2}}\bigg) \ with \ r \ = \ \frac{K}{1-K^{2}}\)

\(\displaystyle Note: \ 0 \ < \ K \ < \ 1, \ If \ K \ = \ 0, \ then \ we \ have \ a \ point \ (at \ the \ origin), \ and \ if \ K \ > \ 1, r \ is \ negative.\)