Multiple choice

Circle $K$ is inscribed in the first quadrant touching the circle $x^{2}+y^{2}=36$ internally. The length of the radius of the circle $K$ is ?

  1. $\dfrac { 6-\sqrt { 2 } }{ 2 } $
  2. $\dfrac { 3-\sqrt { 2 } }{ 2 } $
  3. $3$
  4. $6\left( \sqrt { 2 } -1 \right)$
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D Correct answer
Explanation

The circle x^2 + y^2 = 36 has radius 6 and center (0,0). A circle K with radius r in the first quadrant touching the axes and the circle internally has center (r, r). The distance from (0,0) to (r,r) is r*sqrt(2). The condition for internal tangency is r*sqrt(2) + r = 6. Thus r(sqrt(2)+1) = 6, so r = 6 / (sqrt(2)+1) = 6(sqrt(2)-1).

AI explanation

Because circle K is inscribed in the first quadrant and touches both the x and y axes, its center must be at (r, r) where r is its radius. The larger circle x^2 + y^2 = 36 is centered at the origin (0, 0) and has a radius of 6. Since circle K touches the larger circle internally, the distance between their centers equals the difference of their radii. Therefore, sqrt((r - 0)^2 + (r - 0)^2) = 6 - r, which simplifies to r * sqrt(2) = 6 - r. Solving for r yields r = 6 / (sqrt(2) + 1), which rationalizes to 6(sqrt(2) - 1).

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