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Let \[x^2+y^2=ax+by+c\] be the equation for the circle, contained entirely in the first quadrant, which has radius $1$ and passes through the points $(1,2)$ and $(2, 1).$ Find $a+b+c.$

 Aug 10, 2023
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The problem states that (1, 2) and (2, 1) lie on the circle with radius 1. A line segment that connects these points forms a chord on any circle that fit this criteria. I do not recall the name of the theorem, but there exists a theorem that states that the perpendicular bisector of a chord passes through the center of the circle. Luckily, we have enough information to obtain the equation of the perpendicular bisector.

We can start by finding the slope of the perpendicular bisector. This can be done indirectly by finding the slope between (1, 2) and (2, 1) and then realizing that the slope of its perpendicular line is its opposite reciprocal.

\(m_{\text{chord}} = \frac{1 - 2}{2 - 1} = -1 \\ m_{\perp} = 1\)

In addition to knowing the slope of the perpendicular bisector, we also need a point on the line in order to find its equation. We can find a point by finding the midpoint, M, of the chord's endpoints, which is guaranteed to lie on the perpendicular bisector.

\(M = \left(\frac{1 + 2}{2}, \frac{2 + 1}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}\right)\)

A line can be mathematically represented as \(y = mx + b\) where m is the slope and b is the y-intercept. We need to find b in order to find the equation of the line. Note that I use y_c and x_c to emphasize that this equation represents possible coordinates of the center of this circle.

\(y_c = m_{\perp}x_c + b \\ \frac{3}{2} = 1 * \frac{3}{2} + b \\ b = 0 \\ \therefore y_c = x_c\)

Now, we know that y_c = x_c. The equation of a generic circle is \(\left(x - x_c\right)^2 + \left(y - y_c\right)^2 = r^2\). We know one point on the circle is (1, 2). This means that x = 1 and y = 2. We also know that x_c = y_c and r = 1.

\(\left(1 - x_c\right)^2 + \left(2 - x_c\right)^2 = 1^2 \\ 1 - 2x_c + x_c^2 + 4 - 4x_c + x_c^2 = 1 \\ 2x_c^2 - 6x_c + 4 = 0 \\ 2(x_c - 1)(x_c - 2) = 0 \\ x_{c1} = 1, x_{c2} = 2 \\ y_{c1} = 1, y_{c2} = 2\)

In other words, (1, 1), and (2, 2) are candidates for the center of the circle. This means there are two options. Both circles are contained within the first quadrant, so these are both valid options. Now, we just convert to the desired form and find a + b + c.

\((x - 1)^2 + (y - 1)^2 = 1 \\ x^2 - 2x + 1 + y^2 - 2y + 1 = 1 \\ x^2 + y^2 = 2x + 2y - 1 \\ a + b + c = 2 + 2 - 1 = 3\)

Let's evaluate the other option.

\((x - 2)^2 + (y - 2)^2 = 1 \\ x^2 - 4x + 4 + y^2 - 4y + 4 = 1 \\ x^2 + y^2 = 4x + 4y - 7 \\ a + b + c = 4 + 4 - 7 = 1\)

It looks like there are two answers for a + b + c.

 Aug 11, 2023

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