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A student is given the task of designing a model of a new CD. The equation of the circle representing a disc circumference is given by; $$ x^2+y^2-8x+12y-48=0.$$ Determine the radius of the disc and the center of the circle assuming it is plotted on the xy- plane.

Sylvester
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  • What is your difficulty with this question? – Mark Bennet Sep 12 '13 at 17:10
  • In my attempts to answer this, I rearrange like terms hence $$ x^2-8x+y^2+12y=48.$$ I end up with the equation $$ (x-4)^2-(y+6)^2= 10^2 $$. This resembles the standard equation $$ r^2 = (x-h)^2 +(y-k)^2 $$. So my radius $$ r^2 = 10^2 $$ $$ \therefore r = 10.$$ The center of the cycle if represented by (h,k) = (-4,6). Is this right? Please review. – Sylvester Sep 12 '13 at 17:17
  • The answer given already shows the way to do, which you are trying to follow. However you need $(x-4)^2+(y-6)^2$ (there is a minus sign in your equation. Also note that $x-4=0$ means $x=4$, not $x=-4$ when locating the centre. – Mark Bennet Sep 12 '13 at 17:30

2 Answers2

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Completing the Squares $x^2-2\cdot x\cdot4+4^2+y^2+2\cdot x\cdot6+6^2=48+4^2+6^2$

$$\implies (x-4)^2+(y+6)^2=10^2$$

Now,can you recognize the form?

lab bhattacharjee
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  • @Mark Bennet. Much appreciate for highlighting. It has taken me time to practice and perfect this. I just wasn't sure that I got the concept right, consistency. It is nice that I realise I have picked the concept well and that I made some silly mistakes. Knowing this is a big step for me. I am moving on to Ellipses and hyperbolas. – Sylvester Sep 12 '13 at 19:36
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All you need to do for this question is complete the square to get information from the equation. Remember that the standard form of the equation of a circle is $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the coordinates of the centre, and $r$ is the radius. $$x^2+y^2-8x+12y-48=0$$ $$(x^2-8x+16)-16+(y^2+12y+36)-36-48=0$$ $$(x-4)^2+(y+6)^2-16-36-48=0$$ $$(x-4)^2+(y+6)^2-100=0$$ $$(x-4)^2+(y+6)^2=100$$ $$(x-4)^2+(y+6)^2=10^2$$ The radius is $10$, and the centre of the "disc" is located at $(4,-6)$. I hope that this answers your question.

Anonymous Computer
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