Number of circle(s) touching all the lines $3 x + 4 y - 1 = 0, 4 x - 5 y + 2 = 0 and 6 x + 8 y + 3 = 0$ is

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Solution

Let $L_{1} \equiv 3 x + 4 y - 1 = 0 L_{2} \equiv 4 x - 5 y + 2 = 0 L_{3} \equiv 6 x + 8 y + 3 = 0$

$m_{1} = \frac{- 3}{4}, m_{2} = \frac{4}{5}, m_{3} = \frac{- 3}{4} m_{1} = m_{3} \neq m_{2}$

So, $L_{1}$ and $L_{3}$ are parallel and $L_{2}$ intersects other two lines

Number of circles touchng all the lines are $2$

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