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

Answers (1)
D Divya Prakash Singh

From the figure:

circle center

Let the centre point be O(x,y).

Then the radii of the circle OA,\ OB,\ and\ OC  are equal.

The distance OA:

OA = \sqrt{(x-6)^2+(y+6)^2}

The distance OB:

OB = \sqrt{(x-3)^2+(y+7)^2}

The distance OC:

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

Equating the radii of the same circle.

When equating, OA = OB

\sqrt{(x-6)^2+(y+6)^2}= \sqrt{(x-3)^2+(y+7)^2}  

Squaring both sides and applying a^2-b^2 = (a+b)(a-b)

\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y+7)(y+6-y-7) = 0

\Rightarrow (2x-9)(-3) + (2y+13)(-1) = 0

\Rightarrow -6x+27-2y-13 = 0  or  

\Rightarrow 3x+y -7= 0                                 ...................................(1)

When equating, OA = OC

\sqrt{(x-6)^2+(y+6)^2}= \sqrt{(x-3)^2+(y-3)^2}  

Squaring both sides and applying a^2-b^2 = (a+b)(a-b)

\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y-3)(y+6-y+3) = 0

\Rightarrow (2x-9)(-3) + (2y+3)(9) = 0

\Rightarrow -3x+9y+27 = 0                 ...................................(2)

Now, adding the equations (1) and (2), we get

\Rightarrow 10y = -20

\Rightarrow y = -2.

From equation (1), we get

\Rightarrow 3x-2 = 7

\Rightarrow 3x =9

\Rightarrow x =3

Therefore, the centre of the circle is (3,-2).

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