Suppose $S_{1}$ and $S_{2}$ are two unequal circles; AB and CD are the direct common tangents to these circles. A transverse common tangent PQ cuts AB at R and CD at S. If AB = 10, then RS is

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Solution

The correct option is C
10

Let RB be $α$ and PS be $β$

∴ RP = RA = 10- $α$ $\Rightarrow$ RS = 10 - $α$ + $β$ ...... (1)

Also SQ = SD = 10 - $β$ $\Rightarrow$ RS = 10 - $β$ + $α$ .......(2)

(1) and (2) $\Rightarrow$ $α$ = $β$ , Hence RS = 10

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Suppose S1 and S2 are two unequal circles; AB and CD are the direct common tangents to these circles. A transverse common tangent PQ cuts AB at R and CD at S. If AB = 10, then RS is

The correct option is C 10 Let RB be α and PS be β ∴ RP = RA = 10- α ⇒ RS = 10 - α + β ...... (1) Also SQ = SD = 10 - β ⇒ RS = 10 - β + α .......(2) (1) and (2) ⇒ α = β , Hence RS = 10

Suppose $S_{1}$ and $S_{2}$ are two unequal circles; AB and CD are the direct common tangents to these circles. A transverse common tangent PQ cuts AB at R and CD at S. If AB = 10, then RS is

The correct option is C
10

Let RB be $α$ and PS be $β$

∴ RP = RA = 10- $α$ $\Rightarrow$ RS = 10 - $α$ + $β$ ...... (1)

Also SQ = SD = 10 - $β$ $\Rightarrow$ RS = 10 - $β$ + $α$ .......(2)

(1) and (2) $\Rightarrow$ $α$ = $β$ , Hence RS = 10