Question

Determine an equation of a circle that satisfies the given conditions. Answers must be in standard form. Center (2,8) passes through (-1,7)

Answer 1

The equation of a circle is given by
\[\left( x-h \right)^{2} + \left( y-k \right)^{2} = r ^{2}\] where (h,k) is (x,y) coordinates for the centre of the circle, and r is the radius of the circle. Using this information and the fact that the circle passes a point where x = -1 and y = 7, you can find the value of r. Once you have h, k and r, you have the equation of the circle.

Answer 2

Also, when you're looking for the value of r, remember that it has to be positive because r is really a distance (the radius is the distance from the centre of the circle to its circumference) and because distances have to be positive, r would also have to be evaluated as positive.

Answer 3

Can I use the distance formula to find r?

Answer 4

You could, actually. I didn't even think of that! I was suggesting that you plug in the values of x = -1 and y = 7 into the circle equation (as well as the h and k values) so that you're left with an equation where the only variable is r^2, then solve for r. However, the distance formula should work too!

Answer 5

\(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -1}}\quad ,&{\color{blue}{ 7}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 8}})
\end{array}\qquad
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}
\\ \quad \\
(x-{\color{brown}{ h}})^2+(y-{\color{brown}{ h}})^2=r^2\qquad center\ ({\color{brown}{ h,k}})\quad radius=r\)

Answer 6

hmm sorta forgot the "k" anyhow \(\bf (x-{\color{brown}{ h}})^2+(y-{\color{brown}{ k}})^2=r^2\qquad center\ ({\color{brown}{ h,k}})\quad radius=r\)

Answer 7

for the distance formula is got 7.62 (I rounded it)

Answer 8

@jdoe0001 and @Jesusc

Answer 9

Can we see your working? Also, try not to round it - it's usually preferred to have the answer in exact form, i.e. leave it in surd form.

Answer 10

hmm

Answer 11

\(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -1}}\quad ,&{\color{blue}{ 7}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 8}})
\end{array}\qquad
d = \sqrt{({\color{red}{ 2}}-{\color{red}{ (-1)}})^2 + ({\color{blue}{ 8}}-{\color{blue}{ 7}})^2}=\to ?\)

Answer 12

well... that's not well written.. hamm

Answer 13

actualy is ok anyhow is not 7.62 though

Answer 14

notice that 8-7 = 1

Answer 15

Lol I accidently wrote 8-1 instead of 8-7

Answer 16

But I got 3.16227766

Answer 17

Ok, cool. That's a correct approximation but try to leave it in surd form - the way I like to do this is by using my calculator (sometimes my brain) to work out what's under the surd ONLY. Then, I leave my answer with that number in the surd and that's it.

Answer 18

So when I leave it in surd form that'll be r @Jesusc ?

Answer 19

\(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -1}}\quad ,&{\color{blue}{ 7}})\quad &({\color{red}{ 2}}\quad ,&{\color{blue}{ 8}})
\end{array}
\\ \quad \\

d = \sqrt{({\color{red}{ 2}}-{\color{red}{ (-1)}})^2 + ({\color{blue}{ 8}}-{\color{blue}{ 7}})^2}\to \sqrt{(3)^2+(1)^2}\to \sqrt{10}
\\ \quad \\
\begin{array}{llll}
(x-{\color{brown}{ 2}})^2+(y-{\color{brown}{ 8}})^2=10\\
\bf (x-{\color{brown}{ 2}})^2+(y-{\color{brown}{ 8}})^2=(\sqrt{10})^2
\end{array}
\qquad center\ ({\color{brown}{ 2,8}})\quad radius=\sqrt{10}\)

Answer 20

Yes, the surd form would be r (just as jdoe0001 has illustrated).

Answer 21

Thank you guys for showing me instead of jut giving me the answer! I truly appreciate it. @jdoe0001 @Jesusc

Answer 22

You're welcome :)

Answer 23

yw