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

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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 $(3,-2)$ .

Posted by

Divya Prakash Singh

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