The radical axis of the circles $x^2+y^2+2 g x+2 f y+c=0$ and $2 x^2+2 y^2+3 x+8 y+2 c=0$ touches the circle…

The radical axis of the circles $x^2+y^2+2 g x+2 f y+c=0$ and $2 x^2+2 y^2+3 x+8 y+2 c=0$ touches the circle $x^2+$ $y^2+2 x+2 y+1=0$. Then
  1. $g=\frac{3}{8}$ or $f=1$
  2. $g=\frac{2}{3}$ or $f=3$
  3. $g=\frac{1}{2}$ or $f=1$
  4. $g=\frac{3}{4}$ or $f=2$

Solution

Equation of radical axis is $\left(2 g-\frac{3}{2}\right) x+(2 f-4) y=0$
Since $x$ and $y$ axis touch the circle $x^2+y^2+2 x+2 y+1=0$ $y=\left(\frac{2 g-\frac{3}{2}}{2 f-4}\right) x$ touches the circle and passing through the origin.
$\begin{aligned} & \therefore m=0 \text { or } m=\infty \\ & \Rightarrow 2 g-\frac{3}{2}=0 \Rightarrow g=\frac{3}{4} \text { or } 2 f-4=0 \Rightarrow f=2 .\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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