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 Question from Chen: Find the center radius and equation of a circle in standard form given the following conditions: 1. Tangent to 3x+2y=0 at the point (0,0) and passing through (1,-1) and (6,0)

Hi,

I can show you two ways to approach this problem.

First approach:
The equation of the circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^{2}.$ Use this fact and the fact that the circle passes through $(0,0), (1,-1)$ and $(6,0)$ to write three equations in $h, k$ and $r.$ A careful look at these equations shows that two of them can be easily solved for $h.$ Once you know $h$ you can solve the equations for $k$ and $r.$

Second approach:
The slope of the line $3x + 2y = 0$ is $\large \frac{-3}{2}$ and hence the line trough $(0,0)$ and perpendicular to $3x + 2y = 0$ has slope $\large \frac{2}{3}.$ Write the equation of this line.The center of the circle lies on this line and this fact gives you an equation relating $h$ and $k$ directly. Use the equations developed in the first approach to again solve for $h.$ This and the new equation you just found allow you to find $k.$

I hope this helps,
Penny

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