What is (x^2-14x+49)+(y^2+8y+16)-25=0 ?

The solution to (x^2-14x+49)+(y^2+8y+16)-25=0 is Circle with (a,b)=(7,-4),r=5

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(x^2-14x+49)+(y^2+8y+16)-25=0

(x^{2} - 14 x + 49) + (y^{2} + 8 y + 16) - 25 = 0

(a, b) = (7, - 4), r = 5

Solution steps

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

Rewrite

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 4))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 5

3 x^{2} + 3 y^{2} + 6 x - y = 0

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

x^{2} + y^{2} - 6 x + 4 y - 3 = 0

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

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

What is (x^2-14x+49)+(y^2+8y+16)-25=0 ?
The solution to (x^2-14x+49)+(y^2+8y+16)-25=0 is Circle with (a,b)=(7,-4),r=5