What is 5x^2+5y^2+100x+50y+220=0 ?

The solution to 5x^2+5y^2+100x+50y+220=0 is Circle with (a,b)=(-10,-5),r=9

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5x^2+5y^2+100x+50y+220=0

5 x^{2} + 5 y^{2} + 100 x + 50 y + 220 = 0

(a, b) = (- 10, - 5), r = 9

Solution steps

5 x^{2} + 5 y^{2} + 100 x + 50 y + 220 = 0

Rewrite

5 x^{2} + 5 y^{2} + 100 x + 50 y + 220 = 0

in the form of the standard circle equation

(x - (- 10))^{2} + (y - (- 5))^{2} = 9^{2}

Therefore the circle properties are:

(a, b) = (- 10, - 5), r = 9

x^{2} + y^{2} \leq 4 x

3 x^{2} + 3 y^{2} - 24 x - 24 y - 51 = 0

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

x^{2} + y^{2} \leq 64

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

What is 5x^2+5y^2+100x+50y+220=0 ?
The solution to 5x^2+5y^2+100x+50y+220=0 is Circle with (a,b)=(-10,-5),r=9