Question

A circle has a diameter with endpoints (-7, -1) and (-5, -9).

What is the equation of the circle?
r2 = (x + 6)2 + (y + 5)2
r2 = (x + 4)2 + (y + 1)2
r2 = (x + 6)2 + (y - 1)2
r2 = (x + 4)2 + (y + 5)2

Answer 1

Find length of the diameter using distance formula and half of diameter is radius ... Substitute it in circle equation

Answer 2

Here we know end points given lie on the circle therefore substitute the value to x,y in circle equation ... And u will get 2 equation solve it and find the centre of the circle

Answer 3

\(\bf \text{distance between 2 points}\\
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{((-5)-(-7))^2 + ((-9)-(-1))^2}\\
\textit{that will give you the diameter, half that, is the radius}\\
\textit{to get the center of the circle, get the midpoint}\\
\text{middle point of 2 points }\\
\left(\cfrac{x_2 + x_1}{2},\cfrac{y_2 + y_1}{2} \right) \qquad \qquad \left(\cfrac{(-5) + (-7)}{2},\cfrac{(-9) + (-1)}{2} \right) \implies (h, k)\\
\textit{then plug your values in the circle equation} \qquad
(x-h)^2+(y-k)^2=r^2\)

Answer 4

hmm, got a bit truncated

Answer 5

\(\bf \text{distance between 2 points}\\
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{((-5)-(-7))^2 + ((-9)-(-1))^2}\\
\textit{that will give you the diameter, half that, is the radius}\\
\textit{to get the center of the circle, get the midpoint}\\
\text{middle point of 2 points }\\
\left(\cfrac{x_2 + x_1}{2},\cfrac{y_2 + y_1}{2} \right) \qquad \left(\cfrac{(-5) + (-7)}{2},\cfrac{(-9) + (-1)}{2} \right) \implies (h, k)\\
\textit{then plug your values in the circle equation} \quad
(x-h)^2+(y-k)^2=r^2\)