Question 973·Hard·Math

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\begin{cases} x^{2}+y^{2}=25 \\ y = kx+7 \end{cases}

In the system above, $k$ is a constant. If the system has exactly one real-number solution, what is the positive value of $k$ ?

A circle and a line with exactly one shared point are tangent: the line touches the circle without crossing it. Substitute the line equation into the circle equation to get a quadratic in $x$ . Then set its discriminant to zero, which makes two possible intersections merge into one. A Desmos graph helps you picture the touch, but a line that looks tangent on screen may still cross twice.

Hints

Hint 1
A system’s solutions are points on both graphs. If the line meets the circle exactly once, it touches the circle rather than cutting through it. What must happen to the two intersections a line can usually have?
Hint 2
Use substitution because the line already tells you what $y$ equals. Replace every $y$ in the circle equation with the entire expression $kx+7$ . Keep parentheses around it when you square it.
Hint 3
For $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . It must be zero for exactly one real $x$ -value. After substitution, which expressions are $a$ , $b$ , and $c$ ?

Step-by-step

Substitute and set the discriminant to zero
Step 1
See what one intersection would mean
Set $k=1$ as a trial and type all three lines into Desmos. Click the intersections: $(-4,3)$ and $(-3,4)$ . Those are two shared points. For exactly one, the line must be tangent: it must touch the circle at one point without cutting through it.
Step 2
Substitute for y
At a shared point, the line says $y=kx+7$ . Use substitution: replace the circle’s $y$ with the whole expression $kx+7$ :
$x^2+(kx+7)^2=25.$
Step 3
Make a quadratic in x
Expand the squared parentheses:
$x^2+k^2x^2+14kx+49=25.$
Subtract $25$ :
$x^2+k^2x^2+14kx+24=0.$
Combine the $x^2$ terms:
$(k^2+1)x^2+14kx+24=0.$
Step 4
Use the one-solution condition
Each $x$ in this quadratic gives one $y$ on the line, so count its real $x$ -values. In $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . One shared point requires a zero discriminant: the two roots merge when the square root in the quadratic formula is zero. Here $a=k^2+1$ , $b=14k$ , and $c=24$ , so set it to zero:
$(14k)^2-4(k^2+1)(24)=0.$
Step 5
Simplify the discriminant equation
Square and multiply the coefficients:
$196k^2-96(k^2+1)=0.$
Distribute:
$196k^2-96k^2-96=0.$
Combine like terms:
$100k^2-96=0.$
Step 6
Check the positive root in Desmos
Keep the graph and add $100k_1^2-96\sim0\{k_1>0\}$ . Use $k_1$ because $k=1$ is only the graph’s trial value; $\sim$ tells Desmos to find a fit. The positive root appears under PARAMETERS as $k_1\approx0.9798$ . The restriction excludes the negative root.
Step 7
Find the exact value of k squared
The choices are exact, so solve $100k^2-96=0$ without rounding. Add $96$ :
$100k^2=96.$
Divide by $100$ :
$k^2=\frac{96}{100}.$
Reduce the fraction:
$k^2=\frac{24}{25}.$
Step 8
Take the positive square root
Take the positive square root, as the question requires:
$k=\frac{\sqrt{24}}{5}.$
Since $24=4\times6$ , simplify:
$k=\frac{2\sqrt6}{5}.$
This is the positive slope that makes the line touch the circle at exactly one point. Choice D.

Question 973·Hard·Math

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