Multiple choice

The chord of least length which passes through the point $(2,1)$ of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}- 2\mathrm{x}-4\mathrm{y}-13=0$ is

  1. $2$
  2. $3\sqrt{2}$
  3. $2\sqrt{3}$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The chord of least length passing through a point inside a circle is the one perpendicular to the diameter passing through that point. However, the question asks for the length of the chord passing through (2, 1). The center of the circle x^2 + y^2 - 2x - 4y - 13 = 0 is (1, 2) and the radius is sqrt(1^2 + 2^2 + 13) = sqrt(18) = 3*sqrt(2). The distance from (1, 2) to (2, 1) is sqrt((2-1)^2 + (1-2)^2) = sqrt(2). The half-length of the chord is sqrt(r^2 - d^2) = sqrt(18 - 2) = 4. The full length is 8.

AI explanation

The shortest chord passing through a given internal point is the one perpendicular to the line joining that point to the center. The circle equation x^2 + y^2 - 2x - 4y - 13 = 0 has center (1, 2) and radius sqrt(18). The distance from (1, 2) to the point (2, 1) is sqrt((2-1)^2 + (1-2)^2) = sqrt(2). Using the chord length formula 2 times the square root of (r^2 minus d^2), the length is 2 times the square root of (18 minus 2), which equals 8.

Keep practicing — related questions