Question

The lines $3x-4y+4=0$ and $6x-8y-7=0$ are tangents to the same circle.

• A. radius of the circle $=\frac{3}{4}$
• B. radius of the circle $=\frac{3}{2}$
• C. centre of the circle lies on $12x-16y+1=0$
• D. centre of the circle lies on $12x-16y+31=0$

Answer 1

Answer: (A), (C)

radius of the circle $=\frac{3}{4}$
centre of the circle lies on $12x-16y+1=0$

The given lines being parallel tangents to a circle,

the diameter of the circle is equal to the distance between these lines,

So that the required radius is
$\frac{1}{2}\times \frac{4+\frac{7}{2}}{\sqrt{9+16}}=\frac{1}{2}\times \frac{15}{2}\times \frac{1}{5}=\frac{3}{4}$
The center of the circle lies on the line parallel to the given lines at a distance of $\frac{3}{4}$ from each of them.

So let the equation be $3x-4y+k=0$ ...(1)

then $\frac{k-4}{\sqrt{9+16}}=\pm \frac{3}{4}\Rightarrow k=4\pm \left(\frac{15}{4}\right)\Rightarrow k=\frac{1}{4}$ or $\frac{31}{4}$
For $k=\frac{1}{4},$ distance of (1) from the other line is also $\frac{3}{4}$.

Thus the center lies on the line $12x-16y+1=0$