Question 198·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the $xy$ -plane, the graph of

x^2+y^2-8x-6y=144

is a circle. Line $k$ has slope $\frac{5}{12}$ and is tangent to the circle at the point $(a,b)$ , where $b<0$ . Which choice is the value of $b$ ?

An expanded circle equation and a tangent's slope call for two ideas: finding the circle's center and using a right angle. Graph the whole equation in Desmos; its marked top and bottom give you the center and radius. The radius to a tangent, a line that touches the circle at one point, is perpendicular to that line. Use the sign condition to choose which side of the center holds the point.

Hints

Hint 1
Graph the whole circle equation in Desmos and select the circle. Its marked top and bottom are opposite ends of a vertical diameter, a segment across the circle through its center. What lies halfway between them?
Hint 2
A radius drawn to the point where a tangent touches the circle makes a right angle with the tangent. Perpendicular slopes are negative reciprocals: flip the fraction and change its sign.
Hint 3
The numbers $5$ , $12$ , and $13$ make a right triangle. Match its longest side to the radius, then use $b<0$ to decide whether the point is above or below the center.

Step-by-step

Graph the circle, then use the tangent's slope
Step 1
Find the top and bottom of the circle
Type $x^2+y^2-8x-6y=144$ as written, without adding $y=$ in front. Select the circle. Desmos marks its top at $(4,16)$ and bottom at $(4,-10)$ .
Step 2
Find the circle's center
The center is halfway between those top and bottom points. Type $\frac{16+(-10)}{2}$ ; Desmos shows $3$ . Both points have $x$ -coordinate $4$ , so the center is $(4,3)$ .
Step 3
Find the radius
The top-to-bottom distance is a diameter, so half of it is the radius. Type $\frac{16-(-10)}{2}$ ; Desmos shows $13$ . The tangent point is $13$ units from the center.
Step 4
Turn the tangent's slope into a vertical distance
A slope is vertical change divided by horizontal change. The tangent's slope is $\frac{5}{12}$ , but the radius to its touching point is perpendicular, so its slope is $-\frac{12}{5}$ . That direction has horizontal and vertical changes in the ratio $5:12$ . Since $5$ , $12$ , and $13$ form a right triangle and this radius is $13$ , its vertical change has size $12$ .
Step 5
Choose the point below the center
The center's $y$ -coordinate is $3$ . Because $b<0$ , the tangent point must be $12$ units below the center, not above it. Subtract that vertical change:
$b=3-12=-9.$
So the point's $y$ -coordinate is $-9$ . Choice B.

Question 198·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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