The point of intersection of the circle x<sup>2</sup> + y<sup>2</sup> + 10x - 12y + 51 = 0 and the line 3y + x = 3 is:
From the line, \(x = 3 - 3y\). Substitute into the circle:
\((3 - 3y)^2 + y^2 + 10(3 - 3y) - 12y + 51 = 0\)
\(10y^2 - 60y + 90 = 0 \Rightarrow y^2 - 6y + 9 = 0 \Rightarrow y = 3\) (a repeated root, so the line is tangent).
\(x = 3 - 9 = -6\). The point is \((-6, 3)\).
Answer: (A)
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