What is 4x^2+10x+4y^2-y-4=0 ?

The solution to 4x^2+10x+4y^2-y-4=0 is Circle with (a,b)=(-5/4 , 1/8),r=(sqrt(165))/8

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4x^2+10x+4y^2-y-4=0

4 x^{2} + 10 x + 4 y^{2} - y - 4 = 0

(a, b) = (- \frac{5}{4}, \frac{1}{8}), r = \frac{165}{8}

Solution steps

4 x^{2} + 10 x + 4 y^{2} - y - 4 = 0

Rewrite

4 x^{2} + 10 x + 4 y^{2} - y - 4 = 0

in the form of the standard circle equation

(x - (- \frac{5}{4}))^{2} + (y - \frac{1}{8})^{2} = (\frac{165}{8})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{4}, \frac{1}{8}), r = \frac{165}{8}

3 x^{2} + 3 y^{2} + 8 x + 6 y + 7 = 0

36 x^{2} + 36 y^{2} - 48 x - 36 y + 16 = 0

3 x^{2} + 3 y^{2} - 15 - 0 = 0

x^{2} + 2 x + y^{2} - 8 y + 1 = 0

(x - 6)^{2} + (y - (- 2))^{2} = 25

What is 4x^2+10x+4y^2-y-4=0 ?
The solution to 4x^2+10x+4y^2-y-4=0 is Circle with (a,b)=(-5/4 , 1/8),r=(sqrt(165))/8