Find the centre of a circle passing through the points (6,-6) , (3,-7) and (3,3).
Answers
Answered by
40
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...... :-)
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...... :-)
farhan71:
thank you very much
Answered by
13
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.
: )
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.
: )
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