Question 776537
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The standard form of an equation of a circle centered at *[tex \LARGE (h,\,k)] with radius *[tex \LARGE r] is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2]


Presuming that what you mean by "solution point" is a point that is on the circle, you need to use the distance formula to find the measure of the radius since the distance from the center to any point on the circle is the radius:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{(x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2}]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


However, you can save yourself some work if you use a modified version of the distance formula -- don't take the square root just to square it right away again.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d^2\ =\ r^2\ =\ (x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2]



John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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