Question

Write the equation of the circle that has a diameter with endpoints (6,-1) and (0,-7).

Answer 1

hmmm

what's the radius of it anyway?

Answer 2

This is the whole question, we have to figure out the radius to make the question.

Answer 3

anyway, the raidus will be half the diameter
to get the diameter, get the distance between the endpoints

that is \(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -1}})\quad
% (c,d)
&({\color{red}{ 0}}\quad ,&{\color{blue}{ -7}})
\end{array}\qquad
% distance value
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

and then, find the center of the circle
which happens to be at at the midpoint of the those 2 endpoints
that is \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({\color{red}{ 6}}\quad ,&{\color{blue}{ -1}})\quad
% (c,d)
&({\color{red}{ 0}}\quad ,&{\color{blue}{ -7}})
\end{array}\qquad
% distance value
\left(\cfrac{{\color{red}{ x_2}} + {\color{red}{ x_1}}}{2}\quad ,\quad \cfrac{{\color{blue}{ y_2}} + {\color{blue}{ y_1}}}{2} \right)\)

Answer 4

and once you find the diameter
half of that is the radius

and then use that with the center for the circle's equation
\(\bf (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}}\)

Answer 5

Thanks!

Answer 6

yw

Answer 7

I got a radical for the diameter btw