Question #3e577e4a - SAT Suite Question Bank

ID: 3e577e4a

Modded SAT Question Bank by Abdullah Mallik

A circle in the xy-plane has its center at $(- 4, - 6)$ . Line $k$ is tangent to this circle at the point $(- 7, - 7)$ . What is the slope of line $k$ ?

$negative 3$
$- \frac{1}{3}$
$\frac{1}{3}$
$3$

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Correct Answer: A

Rationale

Choice A is correct. A line that's tangent to a circle is perpendicular to the radius of the circle at the point of tangency. It's given that the circle has its center at $(- 4, - 6)$ and line $k$ is tangent to the circle at the point $(- 7, - 7)$ . The slope of a radius defined by the points $(q, r)$ and $(s, t)$ can be calculated as $\frac{t - r}{s - q}$ . The points $(- 7, - 7)$ and $(- 4, - 6)$ define the radius of the circle at the point of tangency. Therefore, the slope of this radius can be calculated as $\frac{(- 6) - (- 7)}{(- 4) - (- 7)}$ , or $\frac{1}{3}$ . If a line and a radius are perpendicular, the slope of the line must be the negative reciprocal of the slope of the radius. The negative reciprocal of $\frac{1}{3}$ is $negative 3$ . Thus, the slope of line $k$ is $negative 3$ .

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect. This is the slope of the radius of the circle at the point of tangency, not the slope of line $k$ .

Choice D is incorrect and may result from conceptual or calculation errors.

Question Difficulty: Hard