Expert: Paul Klarreich - 5/21/2007

Question
hi i am faced with the following question. (i)Find the coordinates of the centre and the radius of the circle x^2+2x+y^2-4y=4  (ii)by writing x+1=3sin@, show that the parametric equations of this circle are x=-1+3sin@, y=2+3cos@; NB:(@='theta')  (iii)show that the x-coodinates of the points of intersection of this circle with the line x+y=1 are x=-1 +  (3/2) sqrt(2)
              -
ok i tried part (i)and i think it is right. i got out (using completing the square) (x+1)^2+(y-2)^2=(3)^2 centre =(-1,2), radius=3 now after this i am not sure how to proceed.i know the general equations for a circle with centre (a,b) and radius r  are x=a+rcos@, y=b+rsin@ BUT for part (ii) they ask to 'SHOW'..... so i don't think it is as easy as just plugging in the values i got in (i). thank you for your help in advance!!!

Answer
Questioner:   jon
Category:  Calculus
Question:  hi i am faced with the following question.
(i)Find the coordinates of the centre and the radius of the circle

x^2+2x+y^2-4y=4  

(ii)by writing x+1=3sin@, show that the parametric equations of this circle are x=-1+3sin@, y=2+3cos@; NB:(@='theta')  

(iii)show that the x-coodinates of the points of intersection of this circle with the line x+y=1 are
x=-1 +-  (3/2) sqrt(2)

ok i tried part (i)and i think it is right. i got out (using completing the square) (x+1)^2+(y-2)^2=(3)^2 centre =(-1,2), radius=3 now after this i am not sure how to proceed.  

>> Yes, that looks right.


I know the general equations for a circle with centre (a,b) and radius r  are x=a+rcos@, y=b+rsin@ BUT for part (ii) they ask to 'SHOW'..... so i don't think it is as easy as just plugging in the values i got in (i). thank you for your help in advance!!!
........................................
Hi, Jon,

Nothing strikes terror in the hearts of students like the word 'SHOW'.  But all it means is:

"Figure out the ........, and to check your work, the answer should come out to ......"

As to the parametric equations. [I will use 't' for theta: less typing.]

Take your  x + 1 = 3 sin t  and substitute into:

(x+1)^2 + (y-2)^2 = (3)^2

(3 sin t)^2 + (y-2)^2 = (3)^2

Now do some routine algebra and trig:

9 sin^2(t) + (y-2)^2 = 9

            (y-2)^2 = 9 - 9 sin^2(t)

            (y-2)^2 = 9(1 - sin^2(t))

            (y-2)^2 = 9(cos^2(t))

            (y-2)^2 = (3 cos t)^2

                y-2 = 3 cos t

                y   = 2 + 3 cos t


(iii)show that the x-coodinates of the points of intersection of this circle with the line x+y=1 are x=-1 +  (3/2) sqrt(2)

Solve simultaneous equations:

(x+1)^2 + (y-2)^2 = 9    (I)
x + y = 1                (II)
x = 1 - y

Substitute:

(1 - y + 1)^2 + ( y - 2)^2 = 9    (I)

(2 - y)^2 + ( y - 2)^2 = 9    (I)

Expand:

4 - 4y + y^2 + y^2 - 4y + 4 = 9

2y^2 - 8y + 8 = 9

2y^2 - 8y - 1 = 0

Use the quadratic formula:
   8 +- sqrt( 64 - 4(2)(-1) )
y = -------------------------
            4

   8 +- sqrt( 64 + 8 )
y = -------------------
            4

   8 +- sqrt( 72 )
y = ----------------
          4

   8 +- 6 sqrt(2)
y = ---------------
          4

I think you can handle the rest.