Question

Find the centre of a circle passing through the points (6,-6) , (3,-7) and (3,3).

Answer 1

Hiii friend,

Let P(X,Y) be the centre of a circle passing through points A(6,-6) , B(3,-7) and C(3,3).

We know that,

The radius of a circle is always equal.


Therefore,

PA = PB = PC

(PA)² = (PB)² = (PC)²

Then,

PA² = PB² and PB² = PC²


PA² = PB²


=>> (X-6)² + (Y+6)² = (X-3)² + (Y+7)²

=>> X²+(6)² - 2 × X × 6 + Y² +(6)² + 2 × Y × 6 = X²+(3)² - 2 × X × 3 + Y² + (7)² + 2 × Y × 7

=> X²+36 - 12X + Y² +36 + 12Y = X²+9 -6X + Y² +49 + 14Y.


=> X²+Y² - 12X + 12Y +72 = X²+Y² -6X + 14Y + 58


=> X²+Y²-X²-Y² -12X +6X + 12Y -14Y +72 -58 = 0

=> -6X -2Y + 14 = 0

=> 6X + 2Y = 14


And,

PB² = PC²

=> (X-3)² + (Y+7)² = (X-3)² + (Y-3)²

=> X² + (3)² -2 × X × 3 + Y² + (7)² + 2 × Y × 7 = X² + (3)² - 2 × X × 3 + Y² + (3)² - 2 × Y × 3


=> X² + 9 - 6X + Y² + 49 + 14Y = X²+9 - 6X + Y² + 9 - 6Y


=> X²+Y² - 6X + 14Y + 58 = X²+Y² -6X -6Y +18


=> X²+Y² -X² - Y² -6X +6X +14Y +6Y +58 -18 = 0


=> 20Y +40 = 0

=> 20Y = -40

=> Y = -40/20 = -2



And,



6X + 2Y = 14

2(3X + Y) = 14



3X + Y = 7

3X + (-2) = 7


3X -2 = 7

3X = 9

X = 9/3 = 3.


X = 3 and Y = -2


Hence,


The centre of the circle is P(X,Y) = P(3,-2)



HOPE IT WILL HELP YOU...... :-)

Answer 2

Hi ,

Let A ( x1 , y1 ) = ( 6 , - 6 )

B ( x2 , y2 ) = ( 3 , -7 )

C ( x3 , y3 ) = ( 3 , 3 )

AB² = ( x2 - x1 )² + ( y2 - y1 )²

= ( 3 - 6 )² + ( -7 + 6 )²

= 9 + 1

= 10 ---( 1 )

Similarly ,

BC² = 100 ---( 2 )

CA² = 90 ---( 3 )

From ( 1 ) , ( 2 ) and ( 3 ) ,

We conclude that ,

BC² = AB² + AC²

ABC is a right angled Triangle .

Right angle at A .

BC is the hypotenuse.

Now ,

Centre of the circle

= Circumcentre

= Mid point of the B( 3 , -7 ) , C(3 , 3 )

= ( 3 +3/2 , -7 +3 /2 )

= ( 6/2 , -4/2 )

= ( 3 , - 2 )

I hope this helps you.

: )