Let $| z - z_{1} | = 8 a n d | z - z_{2} | = 6$ are two curves as shown which intersect at $P (z_{3}) w h e r e C_{1} (z_{1}) a n d C_{2} (z_{2})$ are centres of the two curves. Given $| z_{1} - z_{2} | = 12, Q (Z_{4}) a n d R (Z_{5})$ are two points on the curves respectively such that $| z_{3} - z_{4} | = | z_{3} - z_{5} |$ . The value of $| z_{4} - z_{5} |^{2}$ is ‘k’, then sum of the digits of ‘k’ is …….

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answer is 7.

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Detailed Solution

$C_{1} Q = C_{1} P = 8 a n d C_{2} P = C_{2} R C_{1} P Q a n d C_{2} P R a r e i s o s c e l e s t r i a n g l e s$

M and N are foot the perpendiculars drawn from the respective centres on the sides PQ and PR
Let PQ = PR = x
QM = PM = PN = NR = x/2
$C_{2} M = a a n d C_{2} N = b i n t r i a n g l e C_{2} P N {(\frac{x}{2})}^{2} + b^{2} = 6^{2} a l s o {(C_{1} C_{2})}^{2} = {(M N)}^{2} + {(C_{1} M - C_{2} N)}^{2} 12^{2} = {(\frac{x}{2} + \frac{x}{2})}^{2} + {(a - b)}^{2} x = \sqrt{130} Q R^{2} = 4 x^{2} = 520$