Question

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)?

Answer 1

Answer: $(x+2)^2+(y+3)^2=9$

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 :-

$(x-h)^2+(y-k)^2=r^2$ , where r is the radius of the circle.....(1)

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

$(-2-(-2))^2+(0-(-3))^2=r^2\\\\\Rightarrow0^2+9=r^2\\\\\Rightarrow r^2=9$

When we put $r^2=9$ and center =(h,k)=(-2,-3)  in (1), we get

$(x-(-2))^2+(y-(-3))^2=9\\\\\Rightarrow(x+2)^2+(y+3)^2=9$

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 :-

$(x+2)^2+(y+3)^2=9$

Answer 2

So the radius is from (-2, -3) to (-2, 0) which is a distance of 3The general form for the equation of a circle is:$(x - h)^2 + (y - k)^2 = r^2$, where the center is (h, k)Plug into the general equation

$(x + 2)^2 + (y + 3)^2 = 3^2 \\ (x + 2)^2 + (y + 3)^2 = 9$