Question 158·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

The line meets the circle at $A$ and $B$, so $\overline{AB}$ joins the two crossings. For points $A(x_A,y_A)$ and $B(x_B,y_B)$, the distance formula is $AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$. Measure between the intersections, not from the center to the line.

Type $x^2+y^2-10x+6y+9=0$ and $3x+4y=14$ on separate Desmos lines. Click both intersections. Desmos shows approximately $(2.728,1.454)$ and $(9.912,-3.934)$. Each point is on the circle and the line, so these are $A$ and $B$.

Enter those rounded coordinates as $A$ and $B$, then type $\operatorname{distance}(A,B)$. Desmos gives about $8.98$. Evaluate each choice in Desmos: their values are about $5.99$, $4.40$, $8.98$, and $13.47$, respectively. The third value matches the measured length, so the exact length of $\overline{AB}$ is $\frac{12\sqrt{14}}{5}$. Choice C.

Graph the circle as written and select it. Desmos marks its top and bottom at $(5,2)$ and $(5,-8)$. Type $\frac{2+(-8)}{2}$ to get the center's $y$-coordinate, $-3$, and $\frac{2-(-8)}{2}$ to get half the vertical diameter, $5$. So the center is $(5,-3)$ and the radius is $5$.

For a line $ax+by=c$ and a point $(h,k)$, the perpendicular distance is $\frac{|ah+bk-c|}{\sqrt{a^2+b^2}}$. The perpendicular direction is $(a,b)$, so this formula measures the shortest path from the center to the line. Add the line in Desmos, then type $\frac{|3(5)+4(-3)-14|}{\sqrt{3^2+4^2}}$. Desmos gives $2.2$, or $\frac{11}{5}$ using its fraction button. This is not $AB$; it is the distance to the line.

A perpendicular from a circle's center to a chord, a segment joining two points on the circle, cuts the chord in half. Call half of $AB$ $h$. The radius, the perpendicular distance, and $h$ form a right triangle, so the Pythagorean theorem gives:

Subtract the distance squared:

Type $\sqrt{5^2-(11/5)^2}$ in Desmos; it gives about $4.49$. To keep the length exact, write both terms with denominator $25$:

Subtract:

Take the positive square root and use $504=36\cdot14$:

Take $\sqrt{36}=6$:

The calculation found half of $AB$, so double it:

Multiply the numerator by $2$:

Type $2\left(\frac{6\sqrt{14}}{5}\right)$ in Desmos to check that it gives about $8.98$, the length between the graphed intersections. The segment has length $\frac{12\sqrt{14}}{5}$. Choice C.