Let Mleft(frac{-7}{2}, frac{-5}{2}right) be t | vedclass

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(B) The equation of the circle is $x^2+y^2+10x+8y-23=0$. The center $O$ of the circle is $(-5, -4)$.Since $M$ is the midpoint of the chord $AB$,the li

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(B) The equation of the circle is $x^2+y^2+10x+8y-23=0$. The center $O$ of the circle is $(-5, -4)$.Since $M$ is the midpoint of the chord $AB$,the line $OM$ is perpendicular to the chord $AB$.The slope of $OM$ is $m_1 = \frac{-4 - (-5/2)}{-5 - (-7/2)} = \frac{-4 + 2.5}{-5 + 3.5} = \frac{-1.5}{-1.5} = 1$.Since $OM \perp AB$,the slope of the chord $AB$ is $m_2 = -1/m_1 = -1$.The equation of the line $AB$ passing through $M\left(\frac{-7}{2}, \frac{-5}{2}\right)$ with slope $-1$ is:$y - (-5/2) = -1(x - (-7/2))$$y + 5/2 = -x - 7/2$$x + y + 6 = 0$To match the form $ax + by + 1 = 0$,we divide by $-6$:$-\frac{1}{6}x - \frac{1}{6}y - 1 = 0$Wait,the given form is $ax + by + 1 = 0$. Let's rewrite $x + y + 6 = 0$ as $\frac{1}{6}x + \frac{1}{6}y + 1 = 0$.Comparing this with $ax + by + 1 = 0$,we get $a = 1/6$ and $b = 1/6$.Therefore,$3a + 3b = 3(1/6) + 3(1/6) = 1/2 + 1/2 = 1$.

Let $M\left(\frac{-7}{2}, \frac{-5}{2}\right)$ be the midpoint of the chord $AB$ of the circle $x^2+y^2+10x+8y-23=0$. If $ax+by+1=0$ is the equation of $AB$,then $3a+3b=$

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