Conic Sections:Cycles

Question

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.

Answer 1

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?

Answer 2

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.