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a circle passes through the points (6,-6) (3, -7) and (3, 3). Find the center and radius of the circle.

 Mar 2, 2022
 #1
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Call the center  ( x, y)

 

We have these 3 equations

(x - 6)^2  + ( y + 6)^2 = r^2       (1)

(x - 3)^2  +  ( y + 7)^2 =  r^2     (2)

(x - 3)^2 +  (y - 3)^2 =  r^2        (3)

 

Subtract the third equation from  the  second and  we  have

 

(y + 7)^2  - (y - 3)^2   =  0        simplify

y^2 + 14y + 49  - y^2 + 6y - 9  =  0

20y + 40  =  0

20y  = - 40

y = -40 / 20 =  - 2

 

Replace this value for y in the  first  two equations and  simplify

( x-6)^2  + ( -2 + 6)^2  = r^2      ⇒ x^2 - 12x + 36  + 16  = r^2     ⇒  x^2 -12x + 52  = r^2    ( 4)

(x - 3)^2  + (-2 + 7)^2 = r^2      ⇒  x^2 - 6x +  9   + 25  = r^2       ⇒  x^2  - 6x + 34  = r^2     (5)

 

Subtract (5) from (4)

-6x + 18  = 0

-6x  = -18

x = -18 / -6  =  3

 

The center is (3, -2)  and  to find the radius  we can  use (1)

(3-6)^2  + (-2 + 6)^2  = r^2

9  +  16  = r^2

25 = r^2       take the positive sqrt

5 = r

 

Here's a graph : https://www.desmos.com/calculator/hwro781e8y

 

cool cool cool

 Mar 2, 2022
 #2
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thanks!

Guest Mar 2, 2022

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