What is 4x^{(2)}+4y^{(2)}+6x+8y-5=0 ?

The solution to 4x^{(2)}+4y^{(2)}+6x+8y-5=0 is Circle with (a,b)=(-3/4 ,-1),r=(3sqrt(5))/4

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4x^{(2)}+4y^{(2)}+6x+8y-5=0

4 x^{(2)} + 4 y^{(2)} + 6 x + 8 y - 5 = 0

(a, b) = (- \frac{3}{4}, - 1), r = \frac{3 5}{4}

Solution steps

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

Rewrite

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, - 1), r = \frac{3 5}{4}

(x + 2)^{2} + (y - 6)^{2} = 10

x^{2} + y^{2} - 6 x + 8 y + 20 = 0

2 x^{2} + 2 y^{2} + 4 x + 8 y = 1

2 x^{2} + 2 y^{2} + 8 x - 20 y - 18 = 0

X^{2} + Y^{2} + 6 X - 8 Y = 11

What is 4x^{(2)}+4y^{(2)}+6x+8y-5=0 ?
The solution to 4x^{(2)}+4y^{(2)}+6x+8y-5=0 is Circle with (a,b)=(-3/4 ,-1),r=(3sqrt(5))/4