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If the lines $3 x-4 y+4=0$ and $6 x-8 y-7=0$ are tangent to circle, then the radius of the circle is
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Verified Answer
The correct answer is:
$\frac{3}{4}$ units
We have, $3 x-4 y+4=0$
and $6 x-8 y-7=0 \Rightarrow 3 x-4 y-\frac{7}{2}=0$
Lines (1) and (2) are parallel to one another and these lines are tangents to the circle.
$\therefore$ Distance between the lines is equal to diameter of circle.
$$
\therefore \mathrm{D}=\frac{\left|4-\left(\frac{-7}{2}\right)\right|}{\sqrt{(3)^2+(-4)^2}}=\frac{\left(4+\frac{7}{2}\right)}{\sqrt{25}}=\frac{15}{2(5)}=\frac{3}{2} \Rightarrow \text { radius }=\frac{3}{4} \text { units }
$$
and $6 x-8 y-7=0 \Rightarrow 3 x-4 y-\frac{7}{2}=0$
Lines (1) and (2) are parallel to one another and these lines are tangents to the circle.
$\therefore$ Distance between the lines is equal to diameter of circle.
$$
\therefore \mathrm{D}=\frac{\left|4-\left(\frac{-7}{2}\right)\right|}{\sqrt{(3)^2+(-4)^2}}=\frac{\left(4+\frac{7}{2}\right)}{\sqrt{25}}=\frac{15}{2(5)}=\frac{3}{2} \Rightarrow \text { radius }=\frac{3}{4} \text { units }
$$
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