The line \(2y = x + 3\) meets the circle \(x^{2} + y^{2} - 2x + 6y - 15 = 0\) at points M and N, where N is in the first quadrant. Find the coordinates of M and N.
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Correct Answer: Option n
Explanation:
Line : \(2y = x + 3\)
\(\therefore y = \frac{1}{2} (x + 3)\)
\(x^{2} + y^{2} - 2x + 6y - 15 = 0\)
\(x^{2} + (\frac{1}{2} (x + 3))^{2} - 2x + 6(\frac{1}{2} (x + 3)) - 15 = 0\)
\(x^{2} + \frac{x^{2}}{4} + \frac{6x}{4} + \frac{9}{4} + 3x + 9 - 15 = 0\)
\(\frac{5x^{2}}{4} + \frac{10x}{4} - \frac{15}{4} = 0\)
\(\equiv 5x^{2} + 10x - 15 = 0\)
\(5x^{2} - 5x + 15x - 15 = 0 \implies 5x(x - 1) + 15(x - 1) = 0\)
\((5x + 15)(x - 1) = 0 \implies \text{x = -3 or 1}\)
When \(x = -3\), \(y = \frac{1}{2} (-3 + 3) = 0\)
When \(x = 1\), \(y = \frac{1}{2} (1 + 3) = 2\)
The coordinates are M(-3, 0) and N(1, 2).
Line : \(2y = x + 3\)
\(\therefore y = \frac{1}{2} (x + 3)\)
\(x^{2} + y^{2} - 2x + 6y - 15 = 0\)
\(x^{2} + (\frac{1}{2} (x + 3))^{2} - 2x + 6(\frac{1}{2} (x + 3)) - 15 = 0\)
\(x^{2} + \frac{x^{2}}{4} + \frac{6x}{4} + \frac{9}{4} + 3x + 9 - 15 = 0\)
\(\frac{5x^{2}}{4} + \frac{10x}{4} - \frac{15}{4} = 0\)
\(\equiv 5x^{2} + 10x - 15 = 0\)
\(5x^{2} - 5x + 15x - 15 = 0 \implies 5x(x - 1) + 15(x - 1) = 0\)
\((5x + 15)(x - 1) = 0 \implies \text{x = -3 or 1}\)
When \(x = -3\), \(y = \frac{1}{2} (-3 + 3) = 0\)
When \(x = 1\), \(y = \frac{1}{2} (1 + 3) = 2\)
The coordinates are M(-3, 0) and N(1, 2).