If the angle of intersection at a point where the two circles with radii 5   c m and 12   c m…

1. $\frac{120}{13}$
2. $\frac{60}{13}$
3. $\frac{13}{5}$
4. $\frac{13}{2}$

Solution

$\because C_{1}P=5$
$C_{2}P=12\&\angle C_{1}P{C}_2={90}^o$
$\therefore C_1{C}_2=13$
$\therefore \Delta C{C}_{2}P$
$\tan \theta =\frac{12}{5}\Rightarrow \sin \theta =\frac{12}{13}$
$C_1{C}_2$ biscets the $PQ$ at $R$ & $C_1{C}_2\perp PQ$
in $\Delta P{C}_{1}R, \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{ }\theta =\frac{PR}{5}$
$\Rightarrow \frac{12}{13}=\frac{PR}{5}\Rightarrow PR=\frac{60}{13}$
& $PQ=2PR=\frac{120}{13}$

Asked in: JEE Main 2019 (12 Apr Shift 1)

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