Show that the point (5, 6) outside the circle: (square of x) + (square of y) = 4?

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Sarvajeet Kumar

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S'=(square of 5) + (square of 6) - 4 = -57.
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The equation of the circle is: x^2 + y^2 = 4 = 2^2. Therefore the radius of the circle is 2 units with its center at (0, 0). So, any point at a distance bigger than 2 units must be outside the circle. Distance between the center (0, 0) and point (5, 6) is: Sq. root of = sq root of (25 + 36) = sq root...
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The equation of the circle is: x^2 + y^2 = 4 = 2^2. Therefore the radius of the circle is 2 units with its center at (0, 0). So, any point at a distance bigger than 2 units must be outside the circle. Distance between the center (0, 0) and point (5, 6) is: Sq. root of [(5-0)^2 + (6-0)^2] = sq root of (25 + 36) = sq root of 61 = 7.8. This is greater than the radius of the circle. So the point (5, 6) is outside the circle. read less
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