JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 17)
Let the circles
C1 : x2 + y2 = 9 and C2 : (x $$-$$ 3)2 + (y $$-$$ 4)2 = 16, intersect at the points X and Y. Suppose that another circle C3 : (x $$-$$ h)2 + (y $$-$$ k)2 = r2 satisfies the following conditions :
(i) Centre of C3 is collinear with the centres of C1 and C2.
(ii) C1 and C2 both lie inside C3 and
(iii) C3 touches C1 at M and C2 at N.
Let the line through X and Y intersect C3 at Z and W, and let a common tangent of C1 and C3 be a tangent to the parabola x2 = 8$$\alpha $$y.
There are some expression given in the List-I whose values are given in List-II below.
Which of the following is the only INCORRECT combination?
C1 : x2 + y2 = 9 and C2 : (x $$-$$ 3)2 + (y $$-$$ 4)2 = 16, intersect at the points X and Y. Suppose that another circle C3 : (x $$-$$ h)2 + (y $$-$$ k)2 = r2 satisfies the following conditions :
(i) Centre of C3 is collinear with the centres of C1 and C2.
(ii) C1 and C2 both lie inside C3 and
(iii) C3 touches C1 at M and C2 at N.
Let the line through X and Y intersect C3 at Z and W, and let a common tangent of C1 and C3 be a tangent to the parabola x2 = 8$$\alpha $$y.
There are some expression given in the List-I whose values are given in List-II below.
Which of the following is the only INCORRECT combination?
(III), (R)
(IV), (S)
(I), (P)
(IV), (U)
Explanation
It is given that, the centres of circles C1, C2 and C3 are co-linear,
$$ \therefore $$ $$\left| {\matrix{ 0 & 0 & 1 \cr 3 & 4 & 1 \cr h & k & 1 \cr } } \right| = 0 \Rightarrow 4h = 3k$$ ....(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.
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