Respuesta :
The equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have:
[tex](x-2)^2+(y+3)^2=16\\\\(x-2)^2+(y-(-3))^2=4^2[tex]
Therefore:
[tex]h=2,\ k=-3\to\text{center:}\ (2,\ -3)\ \text{and}\ r=4[/tex]
The graph matches the circle defined by this equation (x-2)^2 + (y+ 3)^2 = 16 is option D.
What is the equation of the circle with radius r units, centered at (x,y) ?
If a circle O has radius of r units length and that it has got its center positioned at (h, k) point of the coordinate plane,
then its equation is given as:
[tex](x-h)^2 + (y-k)^2 = r^2[/tex]
The given equation of the circle is;
[tex](x-2)^2 + (y+ 3)^2 = 16[/tex]
Where, (h, k) - center
r - radius
here the radius of the circle is 4.
So, the graph matches the circle defined by this equation (x-2)^2 + (y+ 3)^2 = 16 is option D.
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