Question

$\huge{\underline{\underline{\tt{QuesTion:}}}}$

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

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Answer 1

SolutioN :-

Let O ( x , y ) is the centre of circle passing through the points A(6,-6), B(3,-7) and C(3,3).

We know that all radius of circle are same , therefore

OB² = OC²

→ (x-3)² + (y+7)² = (x-3)² + (y-3)²

→ (x-3)² - (x-3)² = (y-3)² - (y+7)²

→ 0 = (2y+4)(3)

→ y = - 2

OC² = AO²

→ (x-3)² + (y-3)² = (x-6)² + (y+6)²

→ (x-3)² - (x-6)² = (-2+6)² - (-2-3)²

→ (x-3)² - (x-6)² = 16 - 25

→ (2x-9)(3) = -9

→ 2x = -3+9

→ x = 3

∴ Centre of circle O ( 3 , -2 )

Answer 2

Given : -

points (6 , -6) ; (3 , -7) ; (3 , 3)

Let the point (6 , -6) be A ; (3 ,-7) be B ; (3 , 3) be C.

Concept to be used : -

Line joining any two points on circumference of circle gives us the chord & intersection of the perpendicular bisectors of any two chords gives us the center of the required circle.

Solution : -

Mid point of AB is =  $(\frac{6+3}{2} , \frac{-6 - 7}{2} ) = (4.5 , -6.5)$

slope of line AB  $m_{AB} = \frac{-7 + 6}{ 3 - 6} =\frac{1}{3}$

equation of ⊥ar bisector of line AB passing through the mid point having slope of -3 (∵m₁m₂ should be -1 for perpendicular lines )can be given by,

$(y + 6.5)= -3 (x - 4.5)$

⇒ $y = - 3x + 7$  _____(1)

Now mid point of BC is = $(\frac{3+3}{2} , \frac{-7+3}{2} ) = (3 , -2)$

slope of BC $m_{BC} =$ ∝ (infinity)

equation of ⊥ar bisector of line BC passing through the mid point with slope zero is given by,

⇒ $y = -2$ ______(2)

point of intersection of lines (1) and (2) is centre of required circle

∴ $O = ( x_{o} , y_{o} ) = ( 3 , - 2 )$

___________________________

More : -

Radius of circle is given by distance between center of circle and any of the  point on the circumference (i.e., A , B ,C ).

⇒ R =  AO = 5 units.

We know the general equation of circle,

$(x-x_{o} )^2 + ( y - y_{o} )^2 = R^2$

⇒ $(x-3 )^2 + ( y + 2 )^2 = 5^2$

∴ $(x-3 )^2 + ( y + 2 )^2 = 25$