If each of the lines 5x + 8y = 13 and 4x - y | vedclass

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(D) The point of intersection of the two lines $5x + 8y = 13$ and $4x - y = 3$ is the center of the circle.Solving $4x - y = 3$ gives $y = 4x - 3$. Su

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$a = 5$ and $b \in (-\infty, 1)$

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(D) The point of intersection of the two lines $5x + 8y = 13$ and $4x - y = 3$ is the center of the circle.Solving $4x - y = 3$ gives $y = 4x - 3$. Substituting into the first equation: $5x + 8(4x - 3) = 13$ $\Rightarrow 5x + 32x - 24 = 13$ $\Rightarrow 37x = 37$ $\Rightarrow x = 1$. Thus $y = 1$. The center is $(1, 1)$.The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,where the center is $(-g, -f)$.Here,$-g = a^2 - 7a + 11 = 1$ $\Rightarrow a^2 - 7a + 10 = 0$ $\Rightarrow (a - 2)(a - 5) = 0$ $\Rightarrow a = 2, 5$.And $-f = a^2 - 6a + 6 = 1$ $\Rightarrow a^2 - 6a + 5 = 0$ $\Rightarrow (a - 1)(a - 5) = 0$ $\Rightarrow a = 1, 5$.For both conditions to hold,$a = 5$.The circle equation becomes $x^2 + y^2 - 2x - 2y + b^3 + 1 = 0$,which is $(x - 1)^2 + (y - 1)^2 = 1 - b^3$.For a real circle,the radius squared must be positive: $1 - b^3 > 0$ $\Rightarrow b^3 < 1$ $\Rightarrow b < 1$.

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

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