Let a circle C touch the lines L_{1}: 4x - 3y | vedclass

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(C) The lines are $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$.Since $L_{1}$ passes through $(-1, 2)$,we have $4(-1) - 3(2) + K_{1} =

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$(x-1)^{2} + (y+2)^{2} = 16$

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(C) The lines are $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$.Since $L_{1}$ passes through $(-1, 2)$,we have $4(-1) - 3(2) + K_{1} = 0$ $\Rightarrow -4 - 6 + K_{1} = 0$ $\Rightarrow K_{1} = 10$.Since $L_{2}$ passes through $(3, -6)$,we have $4(3) - 3(-6) + K_{2} = 0$ $\Rightarrow 12 + 18 + K_{2} = 0$ $\Rightarrow K_{2} = -30$.The distance between the parallel tangents $4x - 3y + 10 = 0$ and $4x - 3y - 30 = 0$ is the diameter $2r = \frac{|10 - (-30)|}{\sqrt{4^{2} + (-3)^{2}}} = \frac{40}{5} = 8$.Thus,the radius $r = 4$.The centre of the circle is the midpoint of the segment connecting $(-1, 2)$ and $(3, -6)$,which is $(\frac{-1+3}{2}, \frac{2-6}{2}) = (1, -2)$.The equation of the circle is $(x - 1)^{2} + (y - (-2))^{2} = r^{2} \Rightarrow (x - 1)^{2} + (y + 2)^{2} = 16$.

Let a circle $C$ touch the lines $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$,where $K_{1}, K_{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(-1, 2)$ and $L_{2}$ at $(3, -6)$,then the equation of the circle $C$ is:

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