Multiple choice

If the length of the tangent from (h, k) to the circle $x^2 \, + \, y^2 \, = \, 16$ is twice the length of the tangent from the same point to the circle $x^2 \, + \, y^2 \, + 2x \, + \, 2y \, = \, 0$, then

  1. $h^2 \, + \, k^2 \, + \, 4h \,+ \, 4k \, + \, 16 \, = \, 0$
  2. $h^2 \, + \, k^2 \, + \, 3h \,+ \, 3k \, = \, 0$
  3. $3h^2 \, + \, 3k^2 \, + \, 8h \,+ \, 8k \, + \, 16 \, = \, 0$
  4. $3h^2 \, + \, 3k^2 \, + \, 4h \,+ \, 4k \, + \, 16 \, = \, 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Length of tangent from (h,k) to x^2+y^2-16=0 is sqrt(h^2+k^2-16). To x^2+y^2+2x+2y=0 is sqrt(h^2+k^2+2h+2k). Given sqrt(h^2+k^2-16) = 2 * sqrt(h^2+k^2+2h+2k). Squaring: h^2+k^2-16 = 4(h^2+k^2+2h+2k) = 4h^2+4k^2+8h+8k. Rearranging: 3h^2+3k^2+8h+8k+16=0.

AI explanation

The length of the tangent from (h, k) to the circle x^2 + y^2 = 16 is given by the formula L1 = root(h^2 + k^2 - 16). For the second circle x^2 + y^2 + 2x + 2y = 0, the tangent length is L2 = root(h^2 + k^2 + 2h + 2k). Since L1 is twice L2, we square both sides to write h^2 + k^2 - 16 = 4(h^2 + k^2 + 2h + 2k). Expanding and rearranging the terms results in 3h^2 + 3k^2 + 8h + 8k + 16 = 0.

Keep practicing — related questions