Verified Solution • 2025–26 Session moderate

QJEEM24Maff31: JEE Main 2024 Mathematics - Class 11 JEEM24Maff31 Solution: Let a circle passing through have its centre at the point . Let …

Let a circle passing through (2,0)(2,0) have its centre at the point (h,k)(\mathrm{h}, \mathrm{k}) . Let (xc,yc)(x_{\mathrm{c}}, y_{\mathrm{c}}) be the point of intersection of the lines 3x+5y=13 x+5 y=1 and (2+c)x+5c2y=1(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1 . If h=limc→1xc and k=limc→1yc, then the equation of the circle is :

Multiple Choice Options

  • A 5x2+5y2−4x−2y−12=05 x^2+5 y^2-4 x-2 y-12=0
  • B 25x2+25y2−20x+2y−60=025 x^2+25 y^2-20 x+2 y-60=0
  • C 25x2+25y2−2x+2y−60=025 x^2+25 y^2-2 x+2 y-60=0
  • D 5x2+5y2−4x+2y−12=05 x^2+5 y^2-4 x+2 y-12=0

Step-by-Step Verified Solution

amp;3x+5y=1amp;(2+c)x+5c2y=1amp;3c2x+5c2y=c2\begin{aligned}& 3 x+5 y=1 \\& (2+c) x+5 c^2 y=1 \\& 3 c^2 x+5 c^2 y=c^2\end{aligned}

Subtracting

amp;(2+c−3c2)x=1−c2amp;xc=1−c22+c−3c2=(1−c)(1+c)(1−c)(3c+2)=c+13c+2amp;y=1−3x5=1−3(c+13c+2)5amp;=3c+2−3c−35(3c+2)amp;yc=−15(3c+2)amp;lim⁡c→1xc=25=hamp;lim⁡c→1yc=−125=k\begin{aligned}& \left(2+c-3 c^2\right) x=1-c^2 \\& x_c=\frac{1-c^2}{2+c-3 c^2}=\frac{(1-c)(1+c)}{(1-c)(3 c+2)}=\frac{c+1}{3 c+2} \\& y=\frac{1-3 x}{5}=\frac{1-3\left(\frac{c+1}{3 c+2}\right)}{5} \\& =\frac{3 c+2-3 c-3}{5(3 c+2)} \\& y_c=\frac{-1}{5(3 c+2)} \\& \lim _{c \rightarrow 1} x_c=\frac{2}{5}=h \\& \lim _{c \rightarrow 1} y_c=\frac{-1}{25}=k\end{aligned}

Equation of circle is

25x2+25y2−20x+2y−60=025 x^2+25 y^2-20 x+2 y-60=0

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