Question 6·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Type $x^2+y^2-8x+6y-11=0$ as written in Desmos, then select the circle. Desmos marks its top and bottom points at $(4,3)$ and $(4,-9)$. You don't need to solve the equation for $y$ to graph the circle.

Those points have the same $x$-coordinate, so the distance between them is a vertical diameter, the distance across the circle through its center. The radius runs from the center to the circle, so it's half that distance. Type $\frac{3-(-9)}{2}$; Desmos prints $6$ for circle A's radius.

A circle's area is $\pi r^2$, so the given area ratio $\frac{25}{9}$ comes from a radius ratio of $\sqrt{\frac{25}{9}}=\frac{5}{3}$. Don't multiply the radius by $\frac{25}{9}$: that factor applies to area, not length. Type $6\sqrt{\frac{25}{9}}$; Desmos prints $10$ for circle B's radius.

Let $O$ be the common center. A tangent touches a circle at one point, and the radius to that point meets the tangent at a right angle. So angles $OQP$ and $ORP$ are each $90^\circ$. In the four-sided figure $OQPR$, the other two angles add to $180^\circ$. Subtract the given outside angle at $P$:

The central angle $QOR$ is at the center, so minor arc $QR$ takes up $\frac{120}{360}$ of circle B's circumference, $2\pi(10)$. To get $k$, the number multiplying $\pi$, type $\frac{120}{360}\cdot2\cdot10$. Desmos shows $6.666\ldots$; click its fraction button to see $\frac{20}{3}$. The minor arc is $\frac{20}{3}\pi$ long, so $k=\frac{20}{3}$. Grid in 20/3.