Let a circle passing through (2,0) have its centre at the point (h, k) . Let (x_c, y_c) be the point of intersection of the lines 3 x+5 y=1 and (2+c) x+5 c^2 y=1 . If h= _ c arrow 1 x_ c and k= _ c arrow 1 y_ c , then

Correct answer: D. 25 x^2+25 y^2-20 x+2 y-60=0 aligned & (2+c) x+5 c^2((1-3 x)/(5))=1 & x=(1-c^2)/(2+c-3 c^2), y=(1-3 x)/(5)=(c-1)/(5(2+c-3 c^2)) & h= _ c arrow 1 ((1-c)(1+c))/((1-c)(2+3 c))=(2)/(5) & K= _ c arrow 1 (c-1)/(-5(c-1)(3 c+2))=-(1)/(25) & Centre((2)/(25),-(1)/(25)) aligned aligned & r= (2-(2)/(5))^2+(0-(1)/(25))^2 = (64)/(25)+(1)/(625) & r= 161 25 & (x-(2)/(5))^2+(y+(1)/(25))^2=(161)/(125) & 25 x^2+25 y^2-20 x+2 y-60=0 aligned

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Circle

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Circle

Let a circle passing through (2,0) have its centre at the point (h,...

Let a circle passing through

(2,0)

have its centre at the point

(h, k)

. Let

\left(x_c, y_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

\mathrm{h}=\lim _{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}

and

\mathrm{k}=\lim _{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}

, then the equation of the circle is :

Options

A $25 x^2+25 y^2-2 x+2 y-60=0$
B $5 x^2+5 y^2-4 x+2 y-12=0$
C $5 x^2+5 y^2-4 x-2 y-12=0$
D $25 x^2+25 y^2-20 x+2 y-60=0$

Correct answer

D. $25 x^2+25 y^2-20 x+2 y-60=0$

Step-by-step solution

\begin{aligned} & (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} \\ & \operatorname{Centre}\left(\frac{2}{25},-\frac{1}{25}\right)\end{aligned}

\begin{aligned} & 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 x^2+25 y^2-20 x+2 y-60=0\end{aligned}

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