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Helga [31]
3 years ago
6

What is the equation of a circle with center (-3, -1) that contains the point (1, 2)?

Mathematics
1 answer:
Leno4ka [110]3 years ago
7 0
So, we know the center is at -3,-1, ok

hmmm what's the radius anyway?   well, we know that there's a point at 1,2 that is on the circle's path...hmmmm what's the distance from the center to that point?  well, is the radius, let's check then.

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -3}}\quad ,&{{ -1}})\quad 
%  (c,d)
&({{ 1}}\quad ,&{{ 2}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
r=\sqrt{[1-(-3)]^2+[2-(-1)]^2}\implies r=\sqrt{(1+3)^2+(2+1)^2}
\\\\\\
r=\sqrt{16+9}\implies r=\sqrt{25}\implies r=5

so, what's the equation of a circle with center at -3, -1 and a radius of 5?

\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{\quad  k}})\qquad 
radius=&{{ r}}\\
&-3&-1&5
\end{array} 
\\\\\\\
[x-(-3)]^2+[y-(-1)]^2=5^2\implies (x+3)^2+(y+1)^2=25
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