What is the standard form of the equation of a circle that has its center at (-2,-3) and passes through the point (-2,0)?

Step-by-step explanation:

The standard form of the equation of a circle that has its center(h,k) and passes through the point (x,y) is given by :-

, where r is the radius of the circle.....(1)

When (h,k)=(-2,-3) and (x,y)=(-2,0), the above equation will become :

When we put and center =(h,k)=(-2,-3)  in (1), we get

Hence, the standard form of the equation of a circle that has its center at (-2,-3) and passes through the point (-2,0) is given by :-

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.