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Problem 6

# 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) $$(2,-1)$$

Expert verified
The center of the circle AB is $$\left(\frac{2}{13}, \frac{3}{13}\right)$$. This corresponds with answer option (b).
See the step by step solution

## 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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