The line x+y+1=0 intersects the circle x^2+y^ | vedclass

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(D) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center of the circle is $C(2, -1)$.Let $M(a, b)$ be the midpoint of the chord $AB$. The lin

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$3$

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(D) The equation of the circle is $x^2+y^2-4x+2y-4=0$. The center of the circle is $C(2, -1)$.Let $M(a, b)$ be the midpoint of the chord $AB$. The line $CM$ is perpendicular to the chord $AB$.The slope of the line $x+y+1=0$ is $m_1 = -1$.Since $CM \perp AB$,the slope of $CM$ is $m_2 = -\frac{1}{m_1} = 1$.The equation of line $CM$ passing through $C(2, -1)$ with slope $1$ is $y - (-1) = 1(x - 2)$,which simplifies to $y = x - 3$ or $x - y = 3$.Since $M(a, b)$ lies on both the line $x+y+1=0$ and the line $x-y=3$,we solve the system:$a+b = -1$$a-b = 3$Adding the two equations gives $2a = 2$,so $a = 1$.Substituting $a=1$ into $a-b=3$,we get $1-b=3$,so $b = -2$.Thus,$a-b = 1 - (-2) = 3$.

The line $x+y+1=0$ intersects the circle $x^2+y^2-4x+2y-4=0$ at the points $A$ and $B$. If $M(a, b)$ is the midpoint of $AB$,then $a-b=$

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