Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 9 April, Shift 1 — Question 14

Let a circle passing through $(2,0)$ have its centre at the point $(\mathrm{h}, \mathrm{k})$ . Let $\left(\mathrm{x}_{\mathrm{c}}, \mathrm{y}_{\mathrm{c}}\right)$ be the point of intersection of the lines $3 x+5 y=1$ and $(2+c) x+$ $5 c^{2} y=1$ . If $h=\lim _{c \rightarrow 1} x_{c}$ and $k=\lim _{c \rightarrow 1} y_{c}$ , then the equation of the circle is :

Option A:
$25 x^{2}+25 y^{2}-20 x+2 y-60=0$
Correct
Option B:
$5 x^{2}+5 y^{2}-4 x-2 y-12=0$
Option C:
$25 x^{2}+25 y^{2}-2 x+2 y-60=0$
Option D:
$5 x^{2}+5 y^{2}-4 x+2 y-12=0$

Answer: A

Step-by-step solution

$(2+c) x+5 c^{2}\left(\frac{1-3 x}{5}\right)=1$

$x=\frac{1-c^{2}}{2+c-3 c^{2}}, y=\frac{1-3 x}{5}=\frac{c-1}{5\left(2+c-3 c^{2}\right)}$

$h=\lim _{c \rightarrow 1} \frac{(1-c)(1+c)}{(1-c)(2+3 c)}=\frac{2}{5}$

$K=\lim _{c \rightarrow 1} \frac{c-1}{-5(c-1)(3 c+2)}=-\frac{1}{25}$

Centre $\left(\frac{2}{25},-\frac{1}{25}\right)$

$r=\sqrt{\left(2-\frac{2}{5}\right)^{2}+\left(0-\frac{1}{25}\right)^{2}}=\sqrt{\frac{64}{25}+\frac{1}{625}}$

$r=\frac{\sqrt{161}}{25}$

$\left(x-\frac{2}{5}\right)^{2}+\left(y+\frac{1}{25}\right)^{2}=\frac{161}{125}$

$\Rightarrow 25 \mathrm{x}^{2}+25 \mathrm{y}^{2}-20 \mathrm{x}+2 \mathrm{y}-60=0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions

Lef f(0, pi) arrow R be a function given by f ( x )= \ beginmatrix ( 8/7 )^frac tan8x tan7x, & 0<x<pi /2 \a-8, & x=pi /2 \\(1+ cotx )^b/a…

Answer: A

Step-by-step solution

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