The circles $S_{1}$ with centre $C_{1} (a_{1}, b_{1})$ and radius $r_{1}$ touches externally the circle $S_{2}$ with centre $C_{2} (a_{2}, b_{2})$ and radius $r_{2}$ . If the tangent at their common point passes through the origin, then

(a_{1}^{2} + a_{2}^{2}) + (b_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

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(a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = r_{1}^{2} - r_{2}^{2}

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{(a_{1}^{2} - b_{2})}^{2} + (a_{2}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

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(a_{1}^{2} - b_{1}^{2}) + (a_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

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Solution

The correct option is B $(a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = r_{1}^{2} - r_{2}^{2}$
Given two circles are
$S_{1} \equiv {(x - a_{1})}^{2} + {(y - b_{1})}_{1}^{2} = r_{1}^{2}$ ...(i)
and $S_{2} \equiv {(x - a_{2})}^{2} + {(y - b_{2})}_{2}^{2} = r_{2}^{2}$ ....(ii)
The equation of the common tangent of these two circles is given by,
$S_{1} - S_{2} = 0$
i.e. $2 x (a_{1} - a_{2}) + 2 y (b_{1} - b_{2}) + (a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2} - r_{2}^{2} = 0$
If this passes through the origin, then
$(a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2} - r_{2}^{2} = 0$
$\Rightarrow (a_{2}^{2} - a_{1}^{2}) + (b_{2}^{2} - b_{1}^{2}) = r_{2}^{2} - r_{1}^{2}$
$\Rightarrow (a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = (r_{1}^{2} - r_{2}^{2})$

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