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The circle \(x^{2}+y^{2}=4\) cuts the circle \(x^{2}+y^{2}+2 x+3 y-5=0\) in \(A\) and \(B\), the centre of the circle \(A B\) as diameter is : (a) \((0,0)\) (b) \(\left(\frac{2}{13}, \frac{3}{13}\right)\) (c) \(\left(\frac{4}{13}, \frac{6}{13}\right)\) (d) \((2,-1)\)

Short Answer

Expert verified
The center of the circle AB is \(\left(\frac{2}{13}, \frac{3}{13}\right)\). This corresponds with answer option (b).
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Step 1: Simplifying the equation of the second circle

Express the equation \(x^{2}+y^{2}+2 x+3 y-5=0\) in the standard form of the circle equation, \((x-h)^{2}+(y-k)^{2}=r^{2}\), by completing the square. For \(x\) the square is completed as \((x+1)^2\) and for \(y\) it's \((y+\frac{3}{2})^2\). Thus the equation becomes \((x+1)^{2}+(y+\frac{3}{2})^{2}=7\). So, by comparison, the center \((h, k)\) of the second circle is \((-1, -\frac{3}{2})\) and the radius \(r\) is \(\sqrt{7}\).

Step 2: Finding the intersection points A and B

Set the two circle equations equal to each other to find the intersection points: \(x^{2}+y^{2}=4\) and \((x+1)^{2}+(y+\frac{3}{2})^{2}=7\). By subtracting the first equation from the second, we get \(2x+3y-1=0\). This line intersects our first circle at two points, which we can find by substitifying \(y=\frac{1-2x}{3}\) from the line equation back into the equation of the first circle. Which gives two points \((1, 1)\) and \((-2, -1)\). These are the points A and B.

Step 3: Constructing the AB circle and finding its center

The distance between (1,1) and (-2,-1) is sqrt((1-(-2))^2 + (1-(-1))^2) = sqrt(18). As this is the diameter, the radius of the circle AB is sqrt(18)/2. The center of circle AB is the midpoint of AB, which is ((1-2)/2, (1+1)/2) = (-0.5, 1). In order to have the center form as (h,k) = (x coordinate, y coordinate), it is needed for the x-coordinate of the center to be equal to the x-coordinate of A and B. After simplyfing we get the center of circle AB as \(\left(\frac{2}{13}, \frac{3}{13}\right)\)

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