If 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…
If 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 \((4 g-3)(f-2)=\)
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-1
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2
Solution
The radical axis of the circles
\(\begin{aligned}
& x^2+y^2+2 g x+2 f y+c=0 \\
& \text { and } \quad 2 x^2+2 y^2+3 x+8 y+2 x=0 \\
& \text { is } (4 g-3) x+(4 f-8) y=0 \quad \ldots (i) \\
\end{aligned}\)
Since, the radical axis (i) touches the circle
\(\begin{aligned}
& x^2+y^2+2 x+2 y+1=0, \text { so } \\
& \frac{|-(4 g-3)-(4 f-8)|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=\sqrt{1+1-1} \\
& \Rightarrow(4 g-3)^2+(4 f-8)^2+2(4 g-3)(4 f-8) \\
& =(4 g-3)^2+(4 f-8)^2 \\
& \Rightarrow 8(4 g-3)(f-2)=0 \\
& \Rightarrow(4 g-3)(f-2=0
\end{aligned}\)
Hence, option (1) is correct.