Super-Hard SAT Geometry Question 6 | Free SAT Prep

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

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Circle A in the $xy$ -plane is represented by the equation $x^2+y^2-8x+6y-11=0$ . Circle B has the same center as circle A, and the area of circle B is $\frac{25}{9}$ times the area of circle A. From a point $P$ outside circle B, two lines are tangent to circle B at points $Q$ and $R$ . If $\angle QPR=60^\circ$ and the length of minor arc $QR$ is $k\pi$ , what is the value of $k$ ?

Your answer

For circle problems combining equations, area ratios, tangents, and arcs, separate the work into stages. First rewrite the equation in standard form to find the original radius. Next, convert the area ratio into a radius ratio by taking its square root. Then use the fact that a radius is perpendicular to a tangent to determine the central angle, and finally apply the arc-length fraction of the circumference.

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Rewrite the circle equation

Complete the square for the $x$ -terms and the $y$ -terms to identify the radius of circle A.

Relate area to radius

Because circle area depends on the square of the radius, take the square root of the given area ratio to obtain the radius ratio.

Use the tangency points

Draw radii from the center to $Q$ and $R$ . Each radius is perpendicular to its tangent, creating two right angles in quadrilateral $OQPR$ .

Use the central angle

After finding $\angle QOR$ , multiply the circumference of circle B by the fraction $\frac{\angle QOR}{360^\circ}$ .

Desmos Guide

Graph circle A

Graph $x^2+y^2-8x+6y-11=0$ . From the graph, identify the center as $(4,-3)$ .

Measure the first radius

Graph $y=-3$ and click its two intersections with the circle. Use half the horizontal distance between those intersections as the radius $a$ of circle A.

Calculate the second radius and central angle

Enter $b=a\sqrt{25/9}$ for the radius of circle B and $t=180-60$ for the central angle subtending minor arc $QR$ .

Calculate the coefficient

Enter $k=(t/360)(2b)$ . The displayed value of $k$ is the requested coefficient of $\pi$ .

Step-by-step Explanation

Find the radius of circle A

Complete the square in both variables:

\begin{aligned} x^2-8x+y^2+6y-11&=0 \\ (x-4)^2-16+(y+3)^2-9-11&=0 \\ (x-4)^2+(y+3)^2&=36 \end{aligned}

Therefore, the radius of circle A is $6$ .

Use the area ratio

Circle area is proportional to the square of the radius. Therefore, the ratio of the radii is the square root of the area ratio:

\frac{r_B}{r_A}=\sqrt{\frac{25}{9}}=\frac{5}{3}

Thus, $r_B=6\left(\frac{5}{3}\right)=10$ .

Find the central angle of the arc

Let $O$ be the common center. A radius drawn to a point of tangency is perpendicular to the tangent, so $\angle OQP$ and $\angle ORP$ are each $90^\circ$ . The angles of quadrilateral $OQPR$ sum to $360^\circ$ , so

\angle QOR=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ.

Calculate the minor arc length

Use the fraction of the circumference determined by the central angle:

\begin{aligned} \text{Arc length}&=\frac{120}{360}\cdot 2\pi(10) \\ &=\frac{20}{3}\pi \end{aligned}

Since the arc length is $k\pi$ , the final answer is $\frac{20}{3}$ .

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