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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: monokekie on January 05, 2009, 09:49:17 am

Title: A circle question
Post by: monokekie on January 05, 2009, 09:49:17 am: hey, could anyone please help me with a question?

it goes like this:

Find the coordinates of the points of intersection of the two curves:

4x^2+4y^2-60x-76y+536=0

and

x^2+y^2-10x-14y+49=0

Thank you sooo much....
Title: Re: A circle question
Post by: /0 on January 05, 2009, 11:04:03 am: $4x^2+4y^2-60x-76y+536=0$ ...[1]

$x^2+y^2-10x-14y+49=0$ ...[2]

Multiply equation [2] by 4:

$4x^2+4y^2-40x-56y+196=0$ ...[2']

Now subtract equation [2'] from equation [1]:

$-20x-20y+340=0$

$x = 17-y$ ...[3]

Now sub [3] back into equation [2]:

$(17-y)^2 + y^2-10(17-y)-14y+49=0$

$y^2-17y+84=0$

$(y-12)(y-7) = 0$

$\therefore y=7,12$

From [2], When $y=7$ , $x^2-10x=0 \Rightarrow x = 0,10$
But substituting $x=0$ and $y=7$ into the first equation, we see that it does not work; it is an extraneous solution.
You can visualise why $x=0$ doesn't work if you draw a graph of the two circles. For the each of $y=7,12$ , you will get two solutions, one of which will be at the intersection point, the other at a point horizontally opposite it. Thus I cannot emphasise more strongly that you check all of your solutions at the end of the problem.

From [2], When $y=12$ , $x^2-10x+25=0 \Rightarrow x = 5$

$\therefore$ Coordinates are $(5,12); (10,7)$
Title: Re: A circle question
Post by: monokekie on January 05, 2009, 11:28:32 am: Thanks sooo much