The circle $S_{1}$ 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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $({a_{1}}^{2} - {a_{2}}^{2}) + ({b_{1}}^{2} - {b_{2}}^{2}) = {r_{1}}^{2} - {r_{2}}^{2}$
The two circles are,

$S_{1} : {(x - a_{1})}^{2} + {(y - b_{1})}^{2} = {r_{1}}^{2}$ ...(1)
$S_{2} : {(x - a_{2})}^{2} + {(y - b_{2})}^{2} = {r_{2}}^{2}$ ...(2)

Subtracting (2) from (1), we get the equation to the common tangent

$S_{1} - S_{2} = 0$

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

If this passes through the origin, then
$({a_{1}}^{2} - {a_{2}}^{2}) + ({b_{1}}^{2} - {b_{2}}^{2}) = {r_{1}}^{2} - {r_{2}}^{2}$

Suggest Corrections

Similar questions

(a_{1}, b_{1}) a n d (a_{2}, b_{2}) i s (a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0

{[{(a_{1} {+ a}_{2} + . . . . a_{n})}^{2} {(b_{1} + b_{2} + . . . . + b_{n})}^{2}]}^{1 / 2} < \sqrt{{a_{1}}^{2} + {b_{1}}^{2} +} \sqrt{{a_{2}}^{2} + {b_{2}}^{2}} + . . . . + \sqrt{{a_{n}}^{2} + {b_{n}}^{2}}

a_{r}, b_{r} (r = 1, 2, ., n)

a_{1} x + b_{1} y + c_{1} z + d_{1} = 0

a_{2} x + b_{2} y + c_{2} z + d_{2} = 0

a_{1} x + b_{1} x + c_{1} x + d_{1} = 0

a_{2} x + b_{2} x + c_{2} x + d_{2} = 0

\frac{x^{2}}{a_{1}^{2}} - \frac{y^{2}}{b_{1}^{2}} = 1

\frac{x^{2}}{a_{2}^{2}} - \frac{y^{2}}{b_{2}^{2}} = 1

Join BYJU'S Learning Program