$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{3}{ h},\stackrel{7}{ k})\qquad \qquad 
radius=\stackrel{6}{ r}
\\\\\\
(x-3)^2+(y-7)^2=6^2\implies (x-3)^2+(y-7)^2=36$

8 0

4 years ago

1 Point The circle below is centered at (-3, 2) and has a radius of 3. What is its equation?

arlik [135]

Answer:

(x + 3)^2 + (y - 2)^2 = 9

Step-by-step explanation:

Equation for a circle: (x – h)^2 + (y – k)^2 = r^2

Step 1: Determine what h, k, and r are

h is the x value of the point: -3

k is the y value of the point: 2

r is the radius of the circle: 3

Step 2: Plug in and solve

(x – h)^2 + (y – k)^2 = r^2

(x – (-3))^2 + (y – (2))^2 = (3)^2

(x + 3)^2 + (y - 2)^2 = 9

Answer: (x + 3)^2 + (y - 2)^2 = 9

8 0

4 years ago

Find the midpoint of the line segment joining the points ​R(​-4,3​) and ​S(​5,6​)

Find the midpoint of the line segment joining the points R(-4,3) and S(5,6)