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...
Question 1 Report
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.
Answer Details
Line \(2y=x+3\) meets circle \(x^{2}+y^{2}-2x+6y-15=0\).
From the line, \(x=2y-3\). Substitute into the circle:
\[(2y-3)^{2}+y^{2}-2(2y-3)+6y-15=0\]\[(4y^{2}-12y+9)+y^{2}-4y+6+6y-15=0\]\[5y^{2}-10y+0=0\;\Rightarrow\;5y(y-2)=0\]So \(y=0\) or \(y=2\).
- \(y=0:\ x=2(0)-3=-3\), giving \((-3,0)\).
- \(y=2:\ x=2(2)-3=1\), giving \((1,2)\).
N is in the first quadrant, so \(N=(1,2)\) and \(M=(-3,0)\).
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