Identify the equation of the circle J with center K(-7,3) and radius 6.

equation of the circle J with center K(-7,3) and radius 6 is=

x^2+y^2+14x-6y+22=0

Step by step explanation:

Equation of circle with given centre (h,k) and radius r is

(x-h)^2+(y-k)^2=r^2

Given centre are K(-7,3) and radius 6.

So equation of Circle=

(x- -7)^2+(y-3)^2=6^2

x^2+14x+49+y^2-6y+9=36

x^2+y^2+14x-6y+22=0 answer

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

it would be:

c.parallelogram.