Question 23·200 Super-Hard SAT Math Questions·Advanced Math

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For the system of equations below, $k$ is a constant. If the system has exactly one real solution $(x,y)$ and $k>0$ , which choice gives the value of $k$ ?

\begin{cases} x^2+y^2=12x \\ y=x+k \end{cases}

A circle and a line can meet twice, touch once, or miss. The cue here is exactly one real solution: the line must touch the circle at one point. Substitute the line equation into the circle equation, then set the resulting quadratic’s discriminant to zero. Desmos can find the positive boundary value, but a graph that merely looks tangent isn’t an exact answer.

Hints

Hint 1
A solution of a system is a point on both graphs. One solution means the line touches the circle once rather than cutting through it at two points.
Hint 2
The line already gives $y$ in terms of $x$ . Replace the whole $y^2$ in the circle equation with $(x+k)^2$ ; each resulting $x$ gives one point on the line.
Hint 3
For a quadratic $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ is zero when its two roots merge into one. What are $a$ , $b$ , and $c$ after substitution?

Step-by-step

Substitute, then count roots
Step 1
See what one solution would look like
Try the positive value $k=2$ to see the graphs. Type $k=2$ , $x^2+y^2=12x$ , and $y=x+k$ in Desmos. They cross twice, near $(0.586,2.586)$ and $(3.414,5.414)$ . Each crossing is a solution, a point on both graphs. We need the value that makes those two crossings merge into one touching point, not the trial value $2$ .
Step 2
Replace y using the line
The line says $y=x+k$ , so substitute $x+k$ for $y$ in the other equation:
$x^2+(x+k)^2=12x.$
Each $x$ that satisfies this equation gives exactly one $y$ on the line. So counting its real $x$ -values counts the system’s solutions.
Step 3
Put the equation in quadratic form
Expand the whole square, including the $2kx$ term:
$x^2+x^2+2kx+k^2=12x.$
Combine the $x^2$ terms:
$2x^2+2kx+k^2=12x.$
Subtract $12x$ :
$2x^2+2kx+k^2-12x=0.$
Group the $x$ terms to get a quadratic in the form $ax^2+bx+c=0$ :
$2x^2+(2k-12)x+k^2=0.$
Step 4
Write the one-root condition
Here $a=2$ , $b=2k-12$ , and $c=k^2$ . The discriminant $b^2-4ac$ is the part under the square root in the quadratic formula. Exactly one real point means the two possible $x$ -values merge, so the discriminant is zero:
$(2k-12)^2-4(2)(k^2)=0.$
A positive discriminant would give two points; a negative one would give none.
Step 5
Find the positive boundary value
Type $(2k_1-12)^2-4(2)k_1^2\sim0\{k_1>0\}$ in Desmos. The subscript makes $k_1$ a value to find rather than the graph’s $k$ slider, and $\sim$ tells Desmos to fit it. Under PARAMETERS, Desmos shows $k_1\approx2.4853$ . The $k_1>0$ restriction selects the value the question allows.
Step 6
Match the decimal to an exact choice
Type the three positive choices into Desmos. It shows $-6+6\sqrt{2}\approx2.4853$ , $6\sqrt{2}\approx8.4853$ , and $6+6\sqrt{2}\approx14.4853$ . Only the first matches the boundary value; the remaining choice is negative, so it breaks $k>0$ . The line touches the circle once when $k=-6+6\sqrt{2}$ . Choice B.

Question 23·200 Super-Hard SAT Math Questions·Advanced Math

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