If two different circles $x^{2} + y^{2} + 2 a x + 2 b y + 1 = 0$ and $x^{2} + y^{2} + 2 b x + 2 a y + 1 = 0$ touches each other, then ${(a + b)}^{2}$ is equal to-

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Solution

The correct option is B $2$
$x^{2} + y^{2} + 2 a x + 2 b y + 1 = 0$
center $= (- a, - b)$
radius $= \sqrt{a^{2} + b^{2} - 1}$
$x^{2} + y^{2} + 2 b x + 1 = 0$
center $= (- b -, - a)$
radius $= \sqrt{a^{2} + b^{2} - 1}$
Since both touch eternally
distance b/w center $= x_{1} + x_{2}$
$\sqrt{(- a + b)^{2} + (- b + a)^{2}} = \sqrt{a^{2} + b^{2} - 1} + \sqrt{a^{2} + b^{2} - 1}$
$\sqrt{2 (a - b)^{2}} = 2 \sqrt{a^{2} + b^{2} - 1}$
$2 a^{2} + 2 b^{2} - 4 a b = 2 a^{2} + 2 b^{2} - 4$
$2 a^{2} + 2 b^{2} + 4 a b = 4$
$a^{2} + b^{2} + 2 a b = 2$
$(a + b)^{2} = 2$

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Similar questions

x^{2} + y^{2} + 2 a x + 2 b y + c = 0

x^{2} + y^{2} + 2 b x + 2 a y + c = 0

x^{2} + y^{2} + 2 a x + c^{2} = 0

x^{2} + y^{2} + 2 b y + c^{2} = 0

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

The two circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch if $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

x^{2} + y^{2} + 2 a x + c = 0

x^{2} + y^{2} + 2 b y + c = 0

\frac{1}{c} =

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x^{2} + y^{2} - 2 b y + c^{2} = 0

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