Question

Equations of circles, the center is (6,-2); passes through (3, 4)

Answer 1

So, we know the center of this circle is at 6, -2, and a point on the circle is at 3,4.  Now, from the center of the circle, to a point on the circle, is how long the radius is, therefore, the distance from 6, -2 to 3,4 is the radius "r", thus,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 6}} &,&{{ -2}}~) % (c,d) &&(~{{3}} &,&{{ 4}}~) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ r=\sqrt{[3-6]^2+[4-(-2)]^2}\implies r=\sqrt{(3-6)^2+(4+2)^2} \\\\\\ r=\sqrt{(-3)^2+6^2}\implies r=\sqrt{45}\\\\ -------------------------------$

$\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad center~~(\stackrel{6}{{{ h}}},\stackrel{-2}{{{ k}}})\qquad \qquad radius=\stackrel{\sqrt{45}}{{{ r}}} \\\\\\\ [x-6]^2+[y-(-2)]^2=(\sqrt{45})^2\implies (x-6)^2+(y+2)^2=45$