The area of a circle having the lines 3x - 4y | vedclass

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(B) The given equations of the parallel tangents are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.To make the coefficients of $x$ and $y$ identical,multipl

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$\frac{9\pi}{16}$

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(B) The given equations of the parallel tangents are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.To make the coefficients of $x$ and $y$ identical,multiply the first equation by $2$:$6x - 8y + 8 = 0$ and $6x - 8y - 7 = 0$.The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Here,the distance between the tangents is the diameter of the circle:$d = \frac{|8 - (-7)|}{\sqrt{6^2 + (-8)^2}} = \frac{15}{\sqrt{36 + 64}} = \frac{15}{10} = \frac{3}{2}$.Since the diameter is $\frac{3}{2}$,the radius $r = \frac{d}{2} = \frac{3}{4}$.The area of the circle is $\pi r^2 = \pi \left(\frac{3}{4}\right)^2 = \frac{9\pi}{16}$ square units.

The area of a circle having the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ as two of its tangents is:

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