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

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asked Nov 23, 2018 176k views

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Stirredo · Answer 1 · 2018-11-30T07:31:46+0000

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,

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

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

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

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