QJEEAdv19Ma82ee: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma82ee Solution: Let the circles 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}

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

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

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

Now, area of

$\Delta MZN = {1 \over 2}(MN)(PZ) = {1 \over 2} \times (12)\left( {{1 \over 2}WZ} \right)$

{ $\because$ MN = 12}

= $3WZ = 3{{24\sqrt 6 } \over 5} = {{72\sqrt 6 } \over 5}$

and area of $\Delta$ ZMW = ${1 \over 2}$ (ZW) (MP)

= ${1 \over 2}\left( {{{24\sqrt 6 } \over 5}} \right)(MG + GP)$

$= {{12\sqrt 6 } \over 5}\left( {3 + {9 \over 5}} \right)$

{ $\because$ MG = 3 and GP = ${{9 \over 5}}$ }

$= {{12\sqrt 6 } \over 5}\left( {{{24} \over 5}} \right)$

= ${{288\sqrt 6 } \over 5}$

$\therefore$ ${{Area\,of\,\Delta MZN} \over {Area\,of\,\Delta ZMW}} = {{{{72\sqrt 6 } \over 5}} \over {{{288\sqrt 6 } \over 5}}} = {5 \over 4}$

$\because$ Common tangent of circles C₁ and C₃ is C₁ $-$ C₃ = 0

$\Rightarrow$ $({x^2} + {y^2} - 9) - \left[ {{{\left( {x - {9 \over 5}} \right)}^2} + {{\left( {y - {{12} \over 5}} \right)}^2} - 36} \right] = 0$

$\Rightarrow$ ${{18} \over 5}x + {{24} \over 5}y + 18 = 0$

$\Rightarrow$ 3x + 4y + 15 = 0 ......(v)

$\because$ Tangent (v) is also touches the parabola x² = 8 $\alpha$ y,

$\therefore$ $- 2\alpha {\left( { - {3 \over 4}} \right)^2} = - {{15} \over 4}$

$\Rightarrow$ $\alpha = {{10} \over 3}$

So combination (iv), (S) is only incorrect.

Hence, option (b) is correct.