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I need this problem from my prep guide answered, it has four answer options and I will provide another picture with the rest of the answer optionsThe subject is pre-Calc

I need this problem from my prep guide answered, it has four answer options and I will provide another picture with the rest of the answer optionsThe subject is class=
I need this problem from my prep guide answered, it has four answer options and I will provide another picture with the rest of the answer optionsThe subject is class=

Answer :

Given,

The equation of circle is,

[tex](x+1)^2+(y+3)^2=16[/tex]

The standard equation of cicle is,

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Where, (h,k) is the center of the circle and r is the radius of the circle.

On compring the standard equation of circle with given equation then,

[tex](h,k)=(-1,-3)[/tex]

Both the coordinates of the center of the circle is negative. Thus, lies in the third quadrant.

Here, the circle of option A has its center at third quadrant.

Hence, option A is correct.

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