Question

find the centre of the circle passing through the points (3,0) ,(2,root5) and (-2root2,-1)?

Answer 1

Answer:

The center of circle is ( 0.16 , 2.1 )

Step-by-step explanation:

Given as :

The points on circle are

A =  (3 , 0)  ,

B = ( 2 , √5 ) ,

C = ( - 2√2 , - 1 )

Let The center of circle O =  (x , y)

we know that, The distance from center on any points on circle are equal

i.e OA = OB = OC

So, From distance formula

OA = OB

$\sqrt{(x - 3)^{2}+(y-2)^{2} }$  = $\sqrt{(x-2)^{2} +(y-\sqrt{5} )^{2} }$

removing the square root

(x - 3)² + (y - 2)²  = (x - 2)² + (y - √5)²

Or, x² - 6 x + 9 + y² - 4 y + 4 = x² - 4 x + 4 + y² - 2√5 y + 5

rearranging the equation'

(x² - x²) + (y² - y²) + (-6 x + 4 x) + (- 4 y + 2√5 y) + 13 - 9 = 0

Or, 0 + 0 - 2 x + (2√5 - 4) y + 4 = 0

Or, - 2 x + (2√5 - 4) y + 4 = 0                       ........1

Again

From distance formula

OA = OC

$\sqrt{(x - 3)^{2}+(y-2)^{2} }$  = $\sqrt{(x+2\sqrt{2} )^{2} +(y + 1)^{2} }$

Removing root both side

x² - 6 x + 9 + y² - 4 y + 4 = x² + 4√ 2 x + 8 + y² + 2 y + 1 =  0

Or, (x² - x²) + (y² - y²) +(- 6 x - 4√ 2 x) + ( - 4 y - 2 y) + 13 - 9 = 0

Or, 0 + 0 - ( 6  + 4√ 2  ) x - 6 y + 4 = 0

Or,   - ( 6  + 4√ 2  ) x - 6 y + 4 = 0                   ........2

Now, Solving the eq 1 and eq 2

x ( - 12 - 24 +12√5 -18√10 +16√2 ) = -24+16+8√5

Or,  x = $\dfrac{2.42}{15}$

i.e  x = 0.16

Put the value of x in eq 2

  - ( 6  + 4√ 2  ) x - 6 y + 4 = 0

i.e    - ( 6  + 4√ 2  ) × 0.16 + 4 = 0 + y

Or, y = 2.1

Hence, The center of circle is ( 0.16 , 2.1 )  Answer

Answer 2

coordinates are (0,0)

Hope it helps