Multiple choice

Consider ${L}{1}:2{x}+3{y}+{p}-3=0$ ${L}{2}:2{x}+3{y}+{p}+3=0$, where ${p}$ is a real number, and ${C}:{x}^{2}+{y}^{2}+6{x}-10{y}+30=0$. STATEMENT 1 : If line $L_{1}$ is a chord of circle $C$, then line $L_{2}$ is not always a diameter of circle $C$. STATEMENT 2 : If line $L_{1}$ is a diameter of circle $C$, then line $L_{2}$ is not a chord of circle $C$.

  1. STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is a correct explanation for STATEMENT 1.

  2. STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is NOT a correct explanation for STATEMENT 1.

  3. STATEMENT 1 is True, STATEMENT 2 is False.

  4. STATEMENT 1 is False, STATEMENT 2 is True.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The circle C has center (-3, 5) and radius sqrt(9 + 25 - 30) = 2. A line 2x + 3y + k = 0 is a diameter if it passes through (-3, 5), meaning 2(-3) + 3(5) + k = 0, so k = -9. For L1, p-3 = -9 implies p = -6. For L2, p+3 = -9 implies p = -12. Since p can vary, L1 being a chord does not force L2 to be a diameter, making Statement 1 true. Statement 2 is false because if L1 is a diameter, L2 is just another line with a different constant term, which could also be a chord or diameter depending on p.

AI explanation

The center of the circle x squared + y squared + 6x - 10y + 30 equals 0 is (-3, 5) and its radius is 2. The parallel lines L1 and L2 have equations with identical slopes and differ only by a constant, meaning they are equidistant from the center. If L1 is a chord, its distance from the center is less than 2, which forces the distance of L2 to be greater than 2, so L2 does not intersect the circle and Statement 1 is true. However, if L1 is a diameter, its distance to the center is 0, which forces L2 to be at a distance of 6 divided by the square root of 13 from the center, allowing it to still be a chord, proving Statement 2 is false.

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