The circles $x^{2} + y^{2} - 4 x + 4 y + 4 = 0$ and $x^{2} + y^{2} - 4 x - 4 y = 0$

Do not intersect

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Are not orthogonal

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Intersect orthogonally

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Concentric

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Solution

The correct option is B Are not orthogonal
$S_{1} = x^{2} + y^{2} + 4 y - 4 x + 4 = 0$
$S_{2} = x^{2} + y^{2} - 4 x - 4 y = 0$
$S_{1} - S_{2} = (x^{2} + y^{2} - 4 x + 4 y + 4) - (x^{2} + y^{2} - 4 x - 4 y)$
$= x^{2} + y^{2} - 4 x + 4 y - 4 - x^{2} - y^{2} + 4 x + 4 y$
$= 8 y - 4$
$\Rightarrow 8 y - 4 = 0 \Rightarrow y = 1 / 2$
from $S_{1} = x^{2} + y^{2} + 4 y - 4 x + 4 = 0$
$C_{1}$ centre (2,-2)
$R_{1}$ Radius = $\sqrt{4 + 4 - 4} = 2$
similarly $S_{2} = X^{2} + y^{2} - 4 x - 4 y = 0$
$C_{2}$ centre (2,2)
$R_{2}$ Radius $= \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$
Distance between centre of circles,
$| C_{1} C_{2} | = \sqrt{(2 - 2)^{2} + (2 + 2)^{2}} = \sqrt{0 + 4^{2}} = \sqrt{4^{2}} = 4$
Now $R_{1} + R_{2} = 2 + 2 \sqrt{2} = 2 (1 + \sqrt{2}) \neq | C_{1} C_{2} |$
$R_{1} + R_{2} >$
The condition for orthogonality $= 2 g g^{'} + 2 f f^{'} = c + c^{'}$
$\Rightarrow 2 (2) (2) + 2 (- 2) (2) = 4 + 0$
$8 - 8 = 0 \neq 4$
So, no circles do not intersect orthogonality,
Option (b) is correct

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