Question

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

Answer 1

hmm do you know what a "perfect square trinomial" is?
sometimes just called a "perfect square"

Answer 2

um..yes

Answer 3

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"?

Answer 4

36, 64

Answer 5

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

Answer 6

or using 64, square root is 8
2 * y * 8 \(\ne 8y\) middle term either

Answer 7

ok

Answer 8

so.. what do you think they might be? you were quite close though

Answer 9

3 and 4

Answer 10

came out a bit off.. lemme fix it quick

Answer 11

its ok. I get the idea thank you so much

Answer 12

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?

Answer 13

and to find any intercepts
say y-intercept, set x = 0

\(\bf x^2+y^2-6x-8y-11=0\implies (0)^2+y^2-6(0)-8y-11=0
\\ \quad \\
y^2-8y-11=0\impliedby \textit{solve for "y"}
\\ \quad \\
\textit{to get the x-intercept, set y=0}
\\ \quad \\
x^2+(0)^2-6x-8(0)-11=0
\\ \quad \\
x^2-6x-11=0\impliedby \textit{solve for "x"}\)