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anonymous
 one year ago
For the equation X^2+y^26x8y11=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
anonymous
 one year ago
For the equation X^2+y^26x8y11=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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm do you know what a "perfect square trinomial" is? sometimes just called a "perfect square"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok.... let us do some grouping first then \(\bf x^2+y^26x8y11=0 \\ \quad \\ (x^26x)+(y^28y)=11 \\ \quad \\ (x^26x+{\color{red}{ \square }}^2)+(y^28y+{\color{red}{ \square }}^2)=11\) any ideas on waht's missing from those groups to get a "perfect square trinomial"?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or using 64, square root is 8 2 * y * 8 \(\ne 8y\) middle term either

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so.. what do you think they might be? you were quite close though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0came out a bit off.. lemme fix it quick

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its ok. I get the idea thank you so much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0keep 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 \(\bf x^2+y^26x8y11=0 \\ \quad \\ (x^26x)+(y^28y)=11 \\ \quad \\ (x^26x+{\color{red}{ 3 }}^2)+(y^28y+{\color{red}{ 4 }}^2){\color{red}{ 3}}^2{\color{red}{ 4}}^2=11 \\ \quad \\ (x3)^2+(y4)^2916=11 \\ \quad \\ (x3)^2+(y4)^2=11+9+16 \\ \quad \\ (x3)^2+(y4)^2=36\implies (x{\color{brown}{ 3 }})^2+(y{\color{blue}{ 4}})^2={\color{purple}{ 6}}^2 \\ \quad \\ \quad \\ (x{\color{brown}{ h}})^2+(y{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}}\) see the center and radius now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and to find any intercepts say yintercept, set x = 0 \(\bf x^2+y^26x8y11=0\implies (0)^2+y^26(0)8y11=0 \\ \quad \\ y^28y11=0\impliedby \textit{solve for "y"} \\ \quad \\ \textit{to get the xintercept, set y=0} \\ \quad \\ x^2+(0)^26x8(0)11=0 \\ \quad \\ x^26x11=0\impliedby \textit{solve for "x"}\)
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