+0  
 
0
5935
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avatar+253 

Find an equation of the circle that satisfies the given conditions.

Center
(2, 10); passes through (9, 7)
 Jun 12, 2014

Best Answer 

 #1
avatar+128732 
+10

Let's find the radius....this is given by ...

√[(-2-9)^2 + (10-7)^2] = √[(-11)^2 + 3^2' = √(121+9) = √130

So the "formula" for an equation of a circle is given by

(x-h)^2 + (y-k)^2 = r^2     where, h,k is the center and r is the radius  ....    so we have...

(x - (-2))^2 + (y-10)^2 = 130

(x +2)^2 + (y-10)^2 = 130

And that's it !!!.....

 Jun 12, 2014
 #1
avatar+128732 
+10
Best Answer

Let's find the radius....this is given by ...

√[(-2-9)^2 + (10-7)^2] = √[(-11)^2 + 3^2' = √(121+9) = √130

So the "formula" for an equation of a circle is given by

(x-h)^2 + (y-k)^2 = r^2     where, h,k is the center and r is the radius  ....    so we have...

(x - (-2))^2 + (y-10)^2 = 130

(x +2)^2 + (y-10)^2 = 130

And that's it !!!.....

CPhill Jun 12, 2014
 #2
avatar+26379 
+6

Find an equation of the circle that satisfies the given conditions.

Center (2, 10); passes through (9, 7)
 
$$Circle: (x-x_0)^2+(y-y_0)^2=r^2 \qquad r=radius \quad Center(x_0,y_0)$$
$$\\\mbox{(1) Center(\textcolor[rgb]{1,0,0}{ -2}, \textcolor[rgb]{1,0,0}{10}): }\quad (x+\textcolor[rgb]{1,0,0}{2})^2+(y-\textcolor[rgb]{1,0,0}{10})^2=r^2\\
\mbox{(2) Point(\textcolor[rgb]{0,0,1}{9}, \textcolor[rgb]{0,0,1}{7}): } \quad (\textcolor[rgb]{0,0,1}{9}+2)^2+(\textcolor[rgb]{0,0,1}{7}-10)^2=r^2\\\\
\mbox{set (1)=(2): } (x+2)^2+(y-10)^2=(9+2)^2+(7-10)^2\\ \\
(x+2)^2+(y-10)^2=11^2+(-3)^2\\\\
(x+2)^2+(y-10)^2=121+9\\\\
\boxed{(x+2)^2+(y-10)^2=130}$$
 Jun 12, 2014

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