Question

Show that the point (11,2) is the centre of the circle passing through the points (1,2), (3,–4) and (5,–6)

Answer 1

Let equation of general equation of circle is x² + y² + 2gx + 2fy + c = 0 then, centre of circle will be (-g , - f)
Circle passing through three points (1,2) , (3,-4) and (5,-6) .
Put (1,2) in equation of circle.
1² + 2² + 2g(1) + 2f(2) + c = 0
2g + 4f + c + 5 = 0 -----(1)

similarly, put (3,-4)
3² + (-4)² + 2g(3) + 2f(-4) + c = 0
9 + 16 + 6g - 8f + c = 0
6g - 8f + c + 25 = 0 ------(2)

put (5, -6),
5² + (-6)² + 2g(5) + 2f(-6) + c = 0
25 + 36 + 10g - 12f + c = 0
10g - 12f + c + 51 = 0 ------(3)

subtracting equation (1) from (2),
6g - 8f + c + 25 - 2g - 4f - c - 5 = 0
4g - 12f + 20 = 0
g - 3f + 5 = 0 -------(4)

subtracting equation (2) from (3),
10g - 12f + c + 51 - 6g + 8f - c - 25 = 0
4g - 4f + 36 = 0
g - f + 9 = 0 -------(5)

solve equations (4) and (5),
f - 9 -3f + 5 = 0
-2f - 4 = 0 ⇒ f = -2
And g = f - 9 = -11
Put g and f in equation (1),
-22 - 8 + c + 5 = 0
⇒ c = 25

Hence, equation of circle is x² + y² -22x -4x + 25 = 0
Centre of circle , C ≡ (- g, - f) ≡ (11, 2)

Hence, it is proved that (11,2) is centre of circle passing through the points (1,2), (3,-4) and (5,-6)

Answer 2

Let A(1 , 2) , B(3 , -4) and C( 5,-6)

are points on the circle .

Let centre of the circle 'O'

i ) radius OA = Radius OB

= Radius OC

ii )A( 1 , 2 ) , O( x , y )

AO² = ( x - 1 )² + ( y - 2 )²

= x² - 2x + 1 + y² - 4y + 4

= x² + y² - 2x - 4y + 5 ----( 1 )

ii ) B( 3 , -4 ) , O( x , y )

BO² = ( x-3 )² + ( y + 4 )²

= x² - 6x + 9 + y² + 8y + 16

= x² + y² - 6x + 8y + 25 ---( 2 )

iii ) C( 5 , -6 ) , O( x , y )

CO² = ( x - 5 )² + ( y + 6 )²

= x² - 10x + 25 + y² + 12y + 36

= x² + y² - 10x + 12y + 61 ----( 3 )

But ,

iv AO² = BO² [ given ]

=> x²+y²-2x-4y+5 = x²+y²-6x+8y+25

=> -2x+6x-4y-8y = 25 - 5

=> 4x-12y = 20

Divide each term with 4 , we get

=> x - 3y = 5 ----( 4 )

iv ) AO² = CO²

=> x²+y²-2x-4y+5 = x²+y²-10x+12y+61

=> -2x+10x-4y-12y = 61-5

=> 8x - 16y = 56

Divide Each term with 8 , we get

=> x - 2y = 7 --( 5 )

Subtract ( 4 ) from ( 5 ) , we get

y = 2 ,

Substitute y = 2 , in equation ( 5 ),

We get

x - 2 × 2 = 7

x = 7 + 4

=> x = 11

Therefore ,

Centre of the circle = O

= ( x , y ) = ( 11 , 2 )

Hence proved.

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