Discuss the position of the points (1, 2) and (6, 0) with respect to the circle $x^{2} + y^{2} - 4 x + 2 y - 11 = 0$

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Solution

We have $x^{2} + y^{2} - 4 x + 2 y - 11 = 0$ or S = 0 where $S = x^{2} + y^{2} - 4 x + 2 y - 11$
For the point (1, 2) we have $S_{1} = 1^{2} + 2^{2} - 4 \times 1 + 2 \times 2 - 11 < 0$
For the point (6, 0) we have $S_{2} = 6^{2} + 0^{2} - 4 \times 6 + 2 \times 0 - 11 > 0$
Hence the point (1, 2) lies inside the circle and the point (6, 0) lies outside the circle

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