OG详解-OG5 数学2 Q26

Choice C is correct. It’s given that the circle has its center at (-1, 1) and that line t is tangent to this circle at the point (5, -4). Therefore, the points (-1, 1) and (5, -4) are the endpoints of the radius of the circle at the point of tangency. The slope of a line or line segment that contains the points (a, b) and (c, d) can be calculated as d−bc−a​​. Substituting (-1, 1) for (a, b) and (5, -4) for (c, d) in the expression d−bc−a​​yields −4−15−(−1​)​  , or - 5​6​. Thus, the slope of this radius is -5​6​. A line that’s tangent to a circle is perpendicular to the radius of the circle at the point of tangency. It follows that line t is perpendicular to the radius at the point (5, -4), so the slope of line t is the negative reciprocal of the slope of this radius. The negative reciprocal of - 5​6​ is 6​5​. Therefore, the slope of line t is 6​5​. Since the slope of line t is the same between any two points on line t, a point lies on line t if the slope of the line segment connecting the point and (5, -4) is 6​5​ .  Substituting choice C, (10, 2), for (a, b) and (5, -4) for (c, d) in the expression d−bc−a yields −4−25−10​​, or 6​5​. Therefore, the point (10, 2) lies on line t.
Choice A is incorrect. The slope of the line segment connecting (0,6​5​ ）and (5, -4) is  −4−65​5−0​​ , or  - 2625​​ , not ​65​. Choice B is incorrect. The slope of the line segment connecting (4, 7) and (5, -4) is −4−75−4​​, or -11, not 6​5​ . Choice D is incorrect. The slope of the line segment connecting (11, 1) and (5, -4) is 
−4−​15−11​ , or 5​6​, not 6​5​  .