Find radius of he circle whose equation is $x^{2} + y^{2} + 8 x + 4 y + 8 = 0$

2 \sqrt{3}

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3

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4 \sqrt{3}

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4

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Solution

The correct option is A $2 \sqrt{3}$
Rectangle width = d
length of rectangle =2 d
Here d is diagonal of each square
p = 2(l+b)= $18 \sqrt{2}$
$= (2 d + d) 2 = 18 \sqrt{2}$
$= d = 3 \sqrt{2}$
As
$d^{2} = 5^{2} + 5^{2}$
$d^{2} = (3 \sqrt{2})^{2} = 18$
$18 = 25^{2}$
$5^{2}$ =9
5 = 3
p square $= 4 \times 5 = 12$

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Find radius of he circle whose equation is x2+y2+8x+4y+8=0

The correct option is A 2√3Rectangle width = dlength of rectangle =2 dHere d is diagonal of each squarep = 2(l+b)=18√2=(2d+d)2=18√2=d=3√2As d2=52+52d2=(3√2)2=1818=25252=95 = 3p square =4×5=12

Find radius of he circle whose equation is $x^{2} + y^{2} + 8 x + 4 y + 8 = 0$