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Difficulty: Hard

A circle in the xy-plane has its center at $left parenthesis negative 1 comma 1 right parenthesis$ . Line $t$ is tangent to this circle at the point $left parenthesis 5 comma negative 4 right parenthesis$ . Which of the following points also lies on line $t$ ?

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Explanation

Choice C is correct. It’s given that the circle has its center at $left parenthesis negative 1 comma 1 right parenthesis$ and that line $t$ is tangent to this circle at the point $left parenthesis 5 comma negative 4 right parenthesis$ . Therefore, the points $left parenthesis negative 1 comma 1 right parenthesis$ and $left parenthesis 5 comma negative 4 right parenthesis$ 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 $\frac{d - b}{c - a}$ . Substituting $left parenthesis negative 1 comma 1 right parenthesis$ for $(a, b)$ and $left parenthesis 5 comma negative 4 right parenthesis$ for $(c, d)$ in the expression $\frac{d - b}{c - a}$ yields $\frac{- 4 - 1}{5 - (- 1)}$ , or $- \frac{5}{6}$ . Thus, the slope of this radius is $- \frac{5}{6}$ . 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 $left parenthesis 5 comma negative 4 right parenthesis$ , so the slope of line $t$ is the negative reciprocal of the slope of this radius. The negative reciprocal of $- \frac{5}{6}$ is $\frac{6}{5}$ . Therefore, the slope of line $t$ is $\frac{6}{5}$ . 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 $left parenthesis 5 comma negative 4 right parenthesis$ is $\frac{6}{5}$ . Substituting choice C, $left parenthesis 10 comma 2 right parenthesis$ , for $(a, b)$ and $left parenthesis 5 comma negative 4 right parenthesis$ for $(c, d)$ in the expression $\frac{d - b}{c - a}$ yields $\frac{- 4 - 2}{5 - 10}$ , or $\frac{6}{5}$ . Therefore, the point $left parenthesis 10 comma 2 right parenthesis$ lies on line $t$ .

Choice A is incorrect. The slope of the line segment connecting $left parenthesis 0 comma six fifths right parenthesis$ and $left parenthesis 5 comma negative 4 right parenthesis$ is $\frac{- 4 - \frac{6}{5}}{5 - 0}$ , or $- \frac{26}{25}$ , not $\frac{6}{5}$ .

Choice B is incorrect. The slope of the line segment connecting $left parenthesis 4 comma 7 right parenthesis$ and $left parenthesis 5 comma negative 4 right parenthesis$ is $\frac{- 4 - 7}{5 - 4}$ , or $negative 11$ , not $\frac{6}{5}$ .

Choice D is incorrect. The slope of the line segment connecting $left parenthesis 11 comma 1 right parenthesis$ and $left parenthesis 5 comma negative 4 right parenthesis$ is $\frac{- 4 - 1}{5 - 11}$ , or $\frac{5}{6}$ , not $\frac{6}{5}$ .