What is the center (x3)^2+(y+4)^2=25 ?

The center (x3)^2+(y+4)^2=25 is (0,-4)

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center (x3)^2+(y+4)^2=25

center

(x 3)^{2} + (y + 4)^{2} = 25

(0, - 4)

Solution steps

(x \cdot 3)^{2} + (y + 4)^{2} = 25

Rewrite

(x 3)^{2} + (y + 4)^{2} = 25

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{3} ) ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 4), a = \frac{5}{3}, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 4), b = 5, a = \frac{5}{3}

And the center is:

(0, - 4)

\frac{( x - 3 ) ^{2}}{5} + \frac{( y + 2 ) ^{2}}{3} = 1

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

4 x^{2} + 7 y^{2} < 28

\frac{( x + 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1