The equation of the circle, concentric with the circle $2 x^2+2 y^2-6 x+8 y+1=0$ and double of its area is

The equation of the circle, concentric with the circle $2 x^2+2 y^2-6 x+8 y+1=0$ and double of its area is
  1. $2 x^2+2 y^2-6 x+8 y+11=0$
  2. $2 x^2+2 y^2-6 x+8 y-11=0$
  3. $4 x^2+4 y^2-12 x+16 y-21=0$
  4. $4 x^2+4 y^2-12 x+16 y+21=0$

Solution

$\begin{aligned} & 2 x^2+2 y^2-6 x+8 y+1=0 \\ & \Rightarrow x^2+y^2-3 x+4 y+\frac{1}{2}=0 \end{aligned}$
Equation of circle concentric to given circle is $x^2+y^2-3 x+4 y+k=0$
Since area of required circle $\begin{aligned} & =2(\text { area of given circle }) \\ & \Rightarrow \pi r_2^2=2\left(\pi r_1^2\right) \\ & \Rightarrow r_2=\sqrt{2} r_1 \\ & \Rightarrow \sqrt{\left(-\frac{3}{2}\right)^2+2^2-\mathrm{k}}=\sqrt{2} \sqrt{\left(\frac{-3}{2}\right)^2+2^2-\frac{1}{2}} \\ & \Rightarrow \sqrt{\frac{25}{4}-\mathrm{k}}=\sqrt{2} \sqrt{\frac{23}{4}} \end{aligned}$ $\begin{aligned} & \Rightarrow \frac{25}{4}-\mathrm{k}=\frac{23}{2} \\ & \Rightarrow \mathrm{k}=\frac{-21}{4}\end{aligned}$ Hence, the required equation is $\begin{aligned} & x^2+y^2-3 x+4 y-\frac{21}{4}=0 \\ & \Rightarrow 4 x^2+4 y^2-12 x+16 y-21=0 \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 1)

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