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anonymous

  • one year ago

For the equation X^2+y^2-6x-8y-11=0, do the following. A) find the center (h,k) and radius r of the circle. B) graph the circle C) find the intercepts, if any

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  1. jdoe0001
    • one year ago
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    hmm do you know what a "perfect square trinomial" is? sometimes just called a "perfect square"

  2. anonymous
    • one year ago
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    um..yes

  3. jdoe0001
    • one year ago
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    ok.... let us do some grouping first then \(\bf x^2+y^2-6x-8y-11=0 \\ \quad \\ (x^2-6x)+(y^2-8y)=11 \\ \quad \\ (x^2-6x+{\color{red}{ \square }}^2)+(y^2-8y+{\color{red}{ \square }}^2)=11\) any ideas on waht's missing from those groups to get a "perfect square trinomial"?

  4. anonymous
    • one year ago
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    36, 64

  5. jdoe0001
    • one year ago
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    hmm 36 and 64 so... 36 square root is 6 so if we multiply 2 * x * 6, we should get the middle term well 2 * x * 6 \(\ne=6x\) though

  6. jdoe0001
    • one year ago
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    or using 64, square root is 8 2 * y * 8 \(\ne 8y\) middle term either

  7. anonymous
    • one year ago
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    ok

  8. jdoe0001
    • one year ago
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    so.. what do you think they might be? you were quite close though

  9. anonymous
    • one year ago
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    3 and 4

  10. jdoe0001
    • one year ago
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    came out a bit off.. lemme fix it quick

  11. anonymous
    • one year ago
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    its ok. I get the idea thank you so much

  12. jdoe0001
    • one year ago
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    keep in mind that, all we're doing is, borrowing from out good fellow, Mr Zero, 0 so if we ADD \(3^2\ and\ 4^2\) we also have to SUBTRACT \(3^2\ and\ 4^2\) thus \(\bf x^2+y^2-6x-8y-11=0 \\ \quad \\ (x^2-6x)+(y^2-8y)=11 \\ \quad \\ (x^2-6x+{\color{red}{ 3 }}^2)+(y^2-8y+{\color{red}{ 4 }}^2)-{\color{red}{ 3}}^2-{\color{red}{ 4}}^2=11 \\ \quad \\ (x-3)^2+(y-4)^2-9-16=11 \\ \quad \\ (x-3)^2+(y-4)^2=11+9+16 \\ \quad \\ (x-3)^2+(y-4)^2=36\implies (x-{\color{brown}{ 3 }})^2+(y-{\color{blue}{ 4}})^2={\color{purple}{ 6}}^2 \\ \quad \\ \quad \\ (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}}\) see the center and radius now?

  13. jdoe0001
    • one year ago
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    and to find any intercepts say y-intercept, set x = 0 \(\bf x^2+y^2-6x-8y-11=0\implies (0)^2+y^2-6(0)-8y-11=0 \\ \quad \\ y^2-8y-11=0\impliedby \textit{solve for "y"} \\ \quad \\ \textit{to get the x-intercept, set y=0} \\ \quad \\ x^2+(0)^2-6x-8(0)-11=0 \\ \quad \\ x^2-6x-11=0\impliedby \textit{solve for "x"}\)

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