Multiple choice

The coordinates of the middle point of the chord which the circle ${ x }^{ 2 }+{ y }^{ 2 }+4x-2y-3=0$ cuts off on the line $y=x+2$, are

  1. $\displaystyle \left( -\frac { 3 }{ 2 } ,\frac { 1 }{ 2 }  \right) $
  2. $\displaystyle \left( \frac { 3 }{ 2 } ,\frac { 1 }{ 2 }  \right) $
  3. $\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 1 }{ 2 }  \right) $
  4. $\displaystyle \left( \frac { 3 }{ 2 } ,-\frac { 1 }{ 2 }  \right) $
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A Correct answer
Explanation

The midpoint of a chord with equation y=mx+c cut by a circle x^2+y^2+2gx+2fy+c=0 is found by substituting the line into the circle equation or using the formula for the chord with a given midpoint. The result is (-3/2, 1/2).

AI explanation

The center of the circle x squared plus y squared + 4x negative 2y negative 3 = 0 is (negative 2, 1). The midpoint of the chord cut off by the line y = x + 2 is the foot of the perpendicular from the circle's center to this line. The equation of the perpendicular from (negative 2, 1) to the line y = x + 2 (which has a slope of 1) is y negative 1 = negative 1(x + 2). Solving the system of the chord and perpendicular equations gives the intersection point (negative 3/2, 1/2).

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