The radius of the circle which has the lines ${x}+{y}-1=0$ and ${x}+{y}-9=0$ as tangents is
- $\sqrt{2}$
- $2\sqrt{2}$
- $3\sqrt{2}$
- $4\sqrt{2}$
The distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by |C1 - C2| / sqrt(A^2 + B^2). Here, the lines are x + y - 1 = 0 and x + y - 9 = 0. The distance is | -1 - (-9) | / sqrt(1^2 + 1^2) = 8 / sqrt(2) = 4 * sqrt(2). The diameter of the circle is this distance, so the radius is half of that, which is 2 * sqrt(2).
The distance between the two parallel tangent lines x + y - 1 = 0 and x + y - 9 = 0 is found by the distance formula |(-1) - (-9)| divided by the square root of 1 squared plus 1 squared, yielding 8 divided by sqrt(2) or 4 sqrt(2). Since the circle is tangent to both lines, its diameter equals the distance between them, meaning the radius is half of 4 sqrt(2). Therefore, the radius is 2 sqrt(2).