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Find the equation of the circle that passes through the points $(0,2)$ and $(6,6)$. Its center is on the line $x-y =1$.

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First, find the center $(a, b)$. It is equidistant from the two points, so $$ \sqrt{a^2 + (b - 2)^2} = \sqrt{(a - 6)^2 + (b - 6)^2} $$ or $$ a^2 + (b - 2)^2 = (a - 6)^2 + (b - 6)^2. $$ Now, since the center is on the line $x - y = 1$, we know that $a - b = 1$, or $$ a = b + 1. $$ Substitute this into the quadratic equation above and simplify: $$ \begin{align} (b + 1)^2 + (b - 2)^2 &= ((b + 1) - 6)^2 + (b - 6)^2 \\ (b^2 + 2b + 1) + (b^2 - 4b + 4) &= (b^2 - 10b + 25) + (b^2 - 12b + 36) \\ 2b^2 - 2b + 5 &= 2b^2 - 22b + 61 \\ 20b &= 56 \\ b &= \frac{14}{5}. \end{align} $$ Therefore, $$ a = \frac{14}{5} + 1 = \frac{19}{5}. $$ What is the radius $r$? Well the square of the radius is $$ \begin{align} r^2 &= a^2 + (b - 2)^2 \\ &= \left( \frac{19}{5} \right)^2 + \left( \frac{4}{5} \right)^2 \\ &= \frac{377}{25}. \end{align} $$ Putting this together, we can write the equation of the circle as $$ \left( x - \frac{19}{5} \right)^2 + \left( y - \frac{14}{5} \right)^2 = \frac{377}{25} $$ or, by multiplying by $25$ to clear denominators, $$ (5x - 19)^2 + (5y - 14)^2 = 377. $$ Here's a picture.

Circle.

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Hint: Find the equation of the line passing through the two given points. This is a chord of the circle. Using the midpoint formula and the fact that perpendicular lines have slopes that are negative reciprocals of each other, find the equation of the perpendicular bisector of this chord. This new line is guaranteed to pass through the centre of circle. Thus, the intersection of this new line and the given line $x-y=1$ must be the centre. After that, you can find the radius by using the distance formula between the centre and one of the given points.

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HINT:

If $(a,b)$ is the center, as it lies on $x-y=1, a-b=1\implies a=b+1$

Now, the distance between $(b+1,b)$ and $(0,2)=\sqrt{(b+1-0)^2+(b-2)^2}=$ the radius of the circle $=$ the distance between $(b+1,b)$ and $(6,6)=\sqrt{(b+1-6)^2+(b-6)^2}$

Solve for $b$

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