QJEEAdv19Ma2654: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma2654 Solution: Let the circle C 1 : x 2 + y 2 = 9 and C 2 : (x 3) 2 + (y 4) 2…

It is given that, the centres of circles C₁, C₂ and C₃ are co-linear,

$\therefore$ $$\left|\begin{array}{ccc}0& 0& 1\\ 3& 4& 1\\ h& k& 1\end{array}\right|=0\Rightarrow 4h=3k$$ ....(i)

and MN is the length of diameter of circle C₃, so

MN = $3 + \sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}} + 4$

= $3 + 5 + 4 = 12$

So, radius of circle C₃, r = 6 ......(ii)

Since, the circle C₃ touches C₁ at M and C₂ at N, so

|C₁ C₃| = |r $-$ 3|

$\Rightarrow$ $\sqrt {{h^2} + {k^2}} = 3$

$\Rightarrow$ h² + k² = 9 .....(iii)

From Eqs. (i) and (iii), we get

${h^2} + {{16{h^2}} \over 9} = 9 \Rightarrow 25{h^2} = 81$

$\Rightarrow h = + {9 \over 5}$ and $k = + {{12} \over 5}$

So, $2h + k = {{18} \over 5} + {{12} \over 5} = 6$

Now, equation common chord XY of circles C₁ and C₂ is

C₁ $-$ C₂ = 0

$\Rightarrow$ 6x + 8y = 18

$\Rightarrow$ 3x + 4y = 9 ....(iv)

Now, PY² = GY² $-$ GP²

= $9 - {{81} \over {25}} = {{144} \over {25}}$

$\Rightarrow$ PY = ${{12} \over 5}$

$\because$ $XY = 2PY = 2 \times {{12} \over 5} = {{24} \over 5}$

Similarly, equation of ZW is 3x + 4y = 9.

Now, So, length of perpendicular from C₃ to

$ZW = {{\left| {3\left( {{9 \over 5}} \right) + 4\left( {{{12} \over 5}} \right) - 9} \right|} \over 5} = {6 \over 5}$

So, $P{W^2} = {C_3}{W^2} - {C_3}{P^2} = 36 - {{36} \over {25}} = {{864} \over {25}}$

{ $\because$ C₃W = r = 6}

$\Rightarrow$ $PW = {{12\sqrt 6 } \over 5}$

$\because$ $ZW = 2PW = {{24\sqrt 6 } \over 5}$

$\therefore$ ${{length\,of\,ZW} \over {length\,of\,XY}} = \sqrt 6$

So, combination (ii), Q is only correct.

Hence, option (c) is correct.