Question

Which of the following graphs matches the circle defined by this equation?

Answer 1

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have:

$(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$

Answer 2

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:

$(x-h)^2 + (y-k)^2 = r^2$

The given equation of the circle is;

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

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.

Learn more about equation of a circle here:

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