The standard equation of a circle is given by (x-h)² + (y-k)² = r², where (h,k) represents the coordinates of the center of the circle and r represents its radius.

Simplify the expression by changing x-(-3) to x+3 and calculating 4² which is 16. This yields the final equation of the circle.

Find the equation of the circle with centre $(-3{,}2)$ and radius 4.

Answer: $(x+3)^2 + (y-2)^2 = 16$

Step-by-step solution

Step 1: Recall the standard equation of a circle

The standard equation of a circle is given by $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ represents the coordinates of the center of the circle and $r$ represents its radius.

Step 2: Identify the given values

From the problem statement, we are given that the center of the circle is $(-3{,}2)$ and the radius is $4$ . Therefore, we can identify $h = -3$ , $k = 2$ , and $r = 4$ .

Step 3: Substitute the values into the equation

Now, substitute the identified values of $h$ , $k$ , and $r$ into the standard equation of the circle. This gives us $(x-(-3))^2 + (y-2)^2 = 4^2$ .

Step 4: Simplify the equation

Simplify the expression by changing $x-(-3)$ to $x+3$ and calculating $4^2$ which is $16$ . This yields the final equation of the circle.

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Find the equation of the circle with centre (−3,2)(-3{,}2)(−3,2) and radius 4.