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Problem 4

# Find the equation of the circle passing through the points of intersection of $$x^{2}+y^{2}=6$$ and $$x^{2}+y^{2}-6 x+8=0$$ and the point $$(1,1)$$.

Expert verified
The equation of the circle passing through the intersection points and the point $$(1,1)$$ is $$(x - 4/3)^{2} + y^{2} = 4$$.
See the step by step solution

## Step 1: Find the intersection points

Subtract the second equation from the first to find the points of intersection. Which gives $$6x - 8 = 0$$, solving this we get $$x = 4/3$$. Substitute $$x = 4/3$$ in the first equation, we get $$y = \pm \sqrt{6 - (4/3)^2}$$, which yields two points of intersection: $$A(4/3, \sqrt{14}/3)$$ and $$B(4/3, -\sqrt{14}/3)$$.

## Step 2: Form the system of equations

The standard equation of a circle is $$(x-a)^{2} + (y-b)^{2} = r^{2}$$, where $$(a, b)$$ is the center and $$r$$ is the radius. Substitute point A, B and the given point $$(1,1)$$ into this equation, we will get three equations.

## Step 3: Solve the system of equations

Solving these equations simultaneously, will give the values $$a = 4/3$$, $$b = 0$$, and $$r = 2$$. Therefore, the equation of the desired circle is $$(x - 4/3)^{2} + y^{2} = 4$$.

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