Question

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

Answer 1

Answer:

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

Answer 2

Answer:

(x + 7)² + ( y − 3)² = 36

Step-by-step explanation:

The equation of a circle with center (h, k) and radius r is                                (x − h)² + (y − k)² = r².

Define h, k and r using the given values. So, h= −7, k = 3, and r = 6.

Substitute the values into the equation of a circle.

(x − (−7))² + (y −3)²= 62

Simplify.

(x + 7)² + (y − 3)²= 36

Therefore, the equation of the circle J with center K (−7, 3) and radius 6 is (x + 7)² + (y − 3)²= 36.