Question

Write the equation of the circle containing the points J(-6, 0), K(-3, 3), and L(0, 0). Show all work to receive credit.

Answer 1

The slope of JK is 1.

The center of the circle is (-3,0)

Which means the radius is the distance between (-3,0) and (-6,0)


D=sqrt of (-6(-3))^2+(0-0)^2
=sqrt of -3^2
=sqrt of 9
r=3

Answer 2

What am I missing?

Answer 3

the equation of a circle with center at (x1,y1) is (x-x1)^2 + (y-y1)^2 = r^2
you have three points and three unknowns
find x1,y1 and r

Answer 4

the equation of a circle with radius 3 and center at (-3,0) is
(x+3)^2+y^2 = 9

Answer 5

okay

so what do i do after (x-3)^2+y^2=9? i mean how do i solve that?

Answer 6

you are asked to write the equation of the circle. That IS the equation of the circle. You don't have to solve for anything

Answer 7

its like asking to write the equation of a line passing through the origin and slope 2
y =2x is the equation of that line. you don't have to solve that equation.

Answer 8

btw, the equation is (x+3)^2+y^2 = 9, not (x-3)^2+y^2=9

Answer 9

Oh okay so i just write it?

Answer 10

yup.

Answer 11

Thank You

Answer 12

you are welcome