Let a circle C touch the lines L _ 1 : 4x - 3y + K _ 1 = 0 and L _...

Given,

${L}_{1} = 4x - 3y + {K}_{1} = 0; {L}_{2} = 4x - 3y + {K}_{2} = 0$

Here line ${L}_{1} and {L}_{2}$ are parallel.

So,

Now point $A\left(- 1, 2\right) will satisfy${L}_{1}$so,$- 4 - 3 \times 2 + {K}_{2} = 0 ⇒ {K}_{2} = 10$</p><p>Also point$B\left(3, - 6\right)$will satisfy${L}_{2} = 4x - 3y + {K}_{2}$</p><p>So$4 \times 3 - 3 \times \left(- 6\right) + {K}_{2} = 0 ⇒ {K}_{2} = - 30$</p><p>Now distance between the parallel line will be the diameter</p><p>Diameter$= \left|\dfrac{{K}_{1} - {K}_{2}}{\sqrt{{4}^{2} + {3}^{2}}}\right|$</p><p>$= \left|\dfrac{10 + 30}{5}\right| = 8$</p><p>So radius$\dfrac{8}{2} = 4$</p><p>Now mid-point of$AB$will give us centre of circle by symmetry, so by midpoint formula in$AB$we get centre \left(1, - 2\right)$,

Now equation of circle will be ${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = {4}^{2}$