sat suite question viewer

Choice C is correct. It’s given that the circle has its center at $\left(-1,1\right)$ and that line $t$ is tangent to this circle at the point $\left(5,-4\right)$. Therefore, the points $\left(-1,1\right)$ and $\left(5,-4\right)$ 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 $\left(a,b\right)$ and $\left(c,d\right)$ can be calculated as $\frac{d-b}{c-a}$. Substituting $\left(-1,1\right)$ for $\left(a,b\right)$ and $\left(5,-4\right)$ for $\left(c,d\right)$ in the expression $\frac{d-b}{c-a}$ yields $\frac{-4-1}{5-\left(-1\right)}$ , 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 $\left(5,-4\right)$, 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 $\left(5,-4\right)$ is $\frac{6}{5}$. Substituting choice C, $\left(10,2\right)$, for $\left(a,b\right)$ and $\left(5,-4\right)$ for $\left(c,d\right)$ in the expression $\frac{d-b}{c-a}$ yields $\frac{-4-2}{5-10}$, or $\frac{6}{5}$. Therefore, the point $\left(10,2\right)$ lies on line $t$.

Choice A is incorrect. The slope of the line segment connecting $\left(0,\frac{6}{5}\right)$ and $\left(5,-4\right)$ is $\frac{-4-{\displaystyle \frac{6}{5}}}{5-0}$, or $-\frac{26}{25}$, not $\frac{6}{5}$.

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

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