Question

of the lines 5x + 8y = 13 and 4x - y = 3 contains a diameter of the circle $x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^2 + 1 = 0$, then:

• A. (1) a = 5 and b ∈ (-∞, 1)
• B. (3) a = 5 and b ∈ (-1, 1)
• C. (2) a = 1 and b ∉ (-1, 1)
• D. (4) a = 2 and b ∈ (-∞, 1)

Answer 1

Answer: (B)

(3)

Answer 2

The intersection of the given lines $5x + 8y = 13$ and $4x - y = 3$ yields the center of the circle as $(1, 1)$. Equating this to the center derived from the circle's equation, $(a^2 - 7a + 11, a^2 - 6a + 6)$, leads to two quadratic equations for $a$. The common solution is $a=5$. For the circle to be real, the radius squared condition $g^2 + f^2 - c > 0$ must be satisfied. With $a=5$, this condition simplifies to $1 - b^2 > 0$, which means $b \in (-1, 1)$. Option (3) correctly states these conditions.