JEE Advanced · Mathematics · 12. Circle

Let the circles $C_{1} : x^{2} + y^{2} = 9$ and $C_{2} : {(x - 3)}^{2} + {(y - 4)}^{2} = 16,$ intersect at the points $X$ and $Y .$ Suppose that another circle $C_{3} : {(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ satisfies the following conditions:
$(i)$ centre of $C_{3}$ is collinear with the centres of $C_{1}$ and $C_{2}$
$(i i) C_{1}$ and $C_{2}$ both lie inside $C_{3},$ and
$(i i i) C_{3}$ touches $C_{1}$ at $M$ and $C_{2}$ at $N .$
Let the line through $X$ and $Y$ intersect $C_{3}$ at $Z$ and $W,$ and let a common tangent of $C_{1}$ and $C_{3}$ be a tangent to the parabola $x^{2} = 8 α y .$
There are some expression given in the List- $I$ whose values are given in List- $I I$ below:

List- $I$ List- II

$(I)$ $2 h + k$ $(P)$ $6$

$(I I)$ $\frac{L e n g t h o f Z W}{L e n g t h o f X Y}$ $(Q)$ $\sqrt{6}$

$(I I I)$ $\frac{A r e a o f t r i a n g l e M Z N}{A r e a o f t r i a n g l e Z M W}$ $(R)$ $\frac{5}{4}$

$(I V)$ $α$ $(S)$ $\frac{21}{5}$

$(T)$ $2 \sqrt{6}$

$(U)$ $\frac{10}{3}$

Which of the following is the only INCORRECT combination?

	List- $I$		List- II
$(I)$	$2 h + k$	$(P)$	$6$
$(I I)$	$\frac{L e n g t h o f Z W}{L e n g t h o f X Y}$	$(Q)$	$\sqrt{6}$
$(I I I)$	$\frac{A r e a o f t r i a n g l e M Z N}{A r e a o f t r i a n g l e Z M W}$	$(R)$	$\frac{5}{4}$
$(I V)$	$α$	$(S)$	$\frac{21}{5}$
		$(T)$	$2 \sqrt{6}$
		$(U)$	$\frac{10}{3}$

A $(I I) - (T)$
B $(I) - (S)$
C $(I) - (U)$
D $(I I) - (Q)$

Verified Solution

Answer & Solution

Correct Answer

(D) $(I I) - (Q)$

Step-by-step Solution

Detailed explanation

Given centre of

C_{1}, C_{2}

and

C_{3}

are collinear hence

|\begin{matrix} 0 & 0 & 1 \\ 3 & 4 & 1 \\ h & k & 1 \end{matrix}| = 0

3 k = 4 h

…(i)

\Rightarrow M N

is diameter of

C_{3}

M N = M C_{1} + C_{1} C_{2} + C_{2} N

\begin{matrix} 2 r = r_{1} + C_{1} C_{2} + r_{2} & (∵ M N = 2 r) \end{matrix}

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

r = 6

…(ii)

\Rightarrow

Given

C_{3}

touches

C_{1}

\begin{matrix} M \end{matrix}

\begin{matrix} |C_{1} C_{3}| = |r - 3| \end{matrix}

h^{2} + k^{2} = 9

…(iii)

From (i) and (iii)

h = \pm \frac{9}{5}

and

k = \pm \frac{12}{5}

So centre of

C_{3}

will be

(\frac{9}{5}, \frac{12}{5})

\Rightarrow

Now equation of common chord of

C_{1}

and

C_{2}

will be

C_{1} - C_{2} = 0

6 x + 8 y = 18

Equation of line

X Y

3 x + 4 y = 9

…(iv)

Distance of line

X Y

from origin

C_{1} P = \frac{9}{5}

\Rightarrow

now in

∆ C_{1} P y

{(C_{1} P)}^{2} + {(P Y)}^{2} = {(C_{1} Y)}^{2}

\begin{matrix} \frac{81}{25} + {(P Y)}^{2} = (9) & (∵ C_{1} Y = r_{1} = 3) \end{matrix}

{(P Y)}^{2} = \frac{144}{25} \Rightarrow P Y = \frac{12}{5}

Length of

X Y = 2 P Y = \frac{24}{5}

\Rightarrow

Line

Z W

is line

X Y

Equation of

Z W = 3 x + 4 y = 9

Distance of

C_{3}

from

Z W = |\frac{3 (\frac{9}{5}) + 4 (\frac{12}{5}) - 9}{5}|

Z W = \frac{6}{5}

now

Z W = 2 \sqrt{{(6)}^{2} - {(\frac{6}{5})}^{2}}

Z W = \frac{24 \sqrt{6}}{5}

(I) 2 h + k = 2 \times \frac{9}{5} + \frac{12}{5} = \frac{30}{5} = 6

(I I) \frac{L e n g t h o f Z W}{L e n g t h o f X Y} = \sqrt{6}

(I I I) \frac{A r e a o f ∆ M Z N}{A r e a o f ∆ Z M W} = \frac{\frac{1}{2} \times M N \times P Z}{\frac{1}{2} \times Z W \times M P}

= \begin{matrix} \frac{\frac{1}{2} M N (\frac{1}{2} Z W)}{\frac{1}{2} \times Z W (M G + G P)} & (∴ P Z = \frac{1}{2} Z W) \end{matrix}

= \begin{matrix} \frac{\frac{1}{2} \times 12 \times \frac{12 \sqrt{6}}{5}}{\frac{1}{2} \times \frac{24 \sqrt{8}}{5} (\frac{24}{5})} & (\begin{matrix} ∴ M C_{1} = 3 \\ ∴ C_{1} A = \frac{9}{5} \end{matrix}) \end{matrix}

= \frac{5}{4}

(I V)

Common tangent to

C_{1}

and

C_{3}

is common chord to

C_{1}

and

C_{3}

C_{1} - C_{3} = 0

3 x + 4 y + 15 = 0

Now this line is tangent to parabola

x^{2} = 8 α y

x^{2} = 8 α (\frac{- 3 x - 15}{y})

4 x^{2} + 24 α x + 120 α = 0

Apply

D = 0

(for tangent, it will have repeated roots)

α = \frac{10}{3}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let

f (x) = x + \log_{e} x - x \log_{e} x, x \in (0, \infty)

•

Column 1 contains information about zeros of

f (x)

f' (x)

and

f'' (x)

•

Column 2 contains information about the limiting behaviour of

f (x)

f' (x)

and

f'' (x)

at infinity.

•

Column 3 contains information about increasing-decreasing nature of

f (x)

and

f' (x)

Column 1	Column 2	Column 3
(I) $f (x) = 0$ for some $x \in (1, e^{2})$	(i) $\lim_{x \to \infty} f (x) = 0$	(P) $f$ is increasing in $(0, 1)$
(II) $f^{'} (x) = 0$ for some $x i n (1, e)$	(ii) $\lim_{x \to \infty} f (x) = - \infty$	(Q) $f$ is decreasing in $(e, e^{2})$
(III) $f^{'} (x) = 0$ for some $x \in (0, 1)$	(iii) $\lim_{x \to \infty} f^{'} (x) = - \infty$	(R) $f^{'}$ is increasing in $(0, 1)$
(IV) $f^{''} (x) = 0$ for some $x \in (1, e)$	(iv) $\lim_{x \to \infty} f^{''} (x) = 0$	(S) $f^{'}$ is decreasing in $(e, e^{2})$

JEE Advanced 2017 Hard

Let $f : R \to R$ and $g : R \to R$ be respectively given by $f (x) = |x| + 1$ and $g (x) = x^{2} + 1 .$ Define $h : R \to R$ by $h (x) = \{\begin{matrix} \max \{f (x), g (x)\} & i f & x \leq 0 \\ \min \{f (x), g (x)\} & i f & x > 0 \end{matrix}$
Then number of points at which $h (x)$ is not differentiable is_________

JEE Advanced 2014 Hard

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