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

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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+130466 
+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

2+0 Answers

 #1
avatar+130466 
+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+26396 
+6

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

Center (−2, 10); passes through (9, 7)
 
Circle:(x−x0)2+(y−y0)2=r2r=radiusCenter(x0,y0)
(1) Center(\textcolor[rgb]{1,0,0}{ -2}, \textcolor[rgb]{1,0,0}{10}): (x+2)2+(y−10)2=r2(2) Point(\textcolor[rgb]{0,0,1}{9}, \textcolor[rgb]{0,0,1}{7}): (9+2)2+(7−10)2=r2set (1)=(2): (x+2)2+(y−10)2=(9+2)2+(7−10)2(x+2)2+(y−10)2=112+(−3)2(x+2)2+(y−10)2=121+9(x+2)2+(y−10)2=130
 Jun 12, 2014

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