If the point $P (x, y)$ lie on a circle with the center (3,-2) and radius 3 unit, then prove that $x^{2} + y^{2} - 6 x + 4 y + 4 = 0$

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Solution

Let center of circle $C (3, - 2)$ and point $P (x, y)$ lie on a circle.
Given: radius of circle $C P = 3$
$\Rightarrow \sqrt{(x - 3)^{2} + (y + 2)^{2}} = 3$
Squaring both sides $\Rightarrow (x - 3)^{2} + (y + 2)^{2} = 9$
$\Rightarrow x^{2} + 9 - 6 x + y^{2} + 4 + 4 y = 9$
$\Rightarrow x^{2} + y^{2} - 6 x + 4 y + 4 = 0$

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(1, - 2)

x^{2} + y^{2} = 1

(3, 2)

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

P (x, y)

x^{2} + y^{2} - 6 x - 6 y + 4 = 0

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

x^{2} + y^{2} + 6 x - 1 = 0

x^{2} + y^{2} - 3 y + 2 = 0

x^{2} + y^{2} + x + y - 3 = 0

S_{1} \equiv x^{2} + y^{2} + 4 y - 1 = 0

S_{2} \equiv x^{2} + y^{2} + 6 x + y + 8 = 0

S_{3} \equiv x^{2} + y^{2} - 4 x - 4 y - 37 = 0

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