Question 964·Hard·Math

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In the system of equations

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

$k$ is a positive real constant, and the system has exactly one real solution $(x,y)$ .

What is the value of $k$ ?

A line and a circle can meet at zero, one, or two points. Exactly one meeting point is called tangency. Substitute the line’s expression for $y$ into the circle’s equation, then use the quadratic’s discriminant to find when there is one real solution. A line that crosses the circle gives two solutions, even if the crossings look close together.

Hints

Hint 1
A solution must satisfy both equations. Since the line already tells you what $y$ equals, replace $y$ in the circle’s equation with $kx+7$ . What kind of equation in $x$ do you get?
Hint 2
For a quadratic equation $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . It equals $0$ when the quadratic has exactly one real solution. What are $a$ , $b$ , and $c$ after you collect terms?
Hint 3
Setting the discriminant to $0$ will give you $k^2$ . Taking a square root gives a positive and a negative possibility, so keep the one allowed by $k$ being positive.

Step-by-step

Substitute and set the discriminant to zero
Step 1
Make one equation in $x$
The line says $y=kx+7$ . Replace $y$ in the circle’s equation with that entire expression, so a solution must satisfy:
$x^2+(kx+7)^2=25.$
Step 2
Expand the square
Square the whole expression $kx+7$ , including its middle term:
$x^2+k^2x^2+14kx+49=25.$
Step 3
Put the quadratic in standard form
Combine the $x^2$ terms:
$(1+k^2)x^2+14kx+49=25.$
Subtract $25$ so one side is $0$ :
$(1+k^2)x^2+14kx+24=0.$
Step 4
Use the one-solution condition
Each real $x$ from this quadratic gives one point on the line. The discriminant of $ax^2+bx+c=0$ is $b^2-4ac$ : positive means two real roots, while zero means one. One shared point means the quadratic’s discriminant is $0$ . Here $a=1+k^2$ , $b=14k$ , and $c=24$ , so:
$(14k)^2-4(1+k^2)(24)=0.$
Step 5
Simplify the discriminant
Square $14k$ and multiply $4$ by $24$ :
$196k^2-96(1+k^2)=0.$
Distribute $-96$ :
$196k^2-96-96k^2=0.$
Combine the $k^2$ terms:
$100k^2-96=0.$
Step 6
Find $k^2$
Add $96$ to both sides:
$100k^2=96.$
Divide by $100$ :
$k^2=\frac{96}{100}.$
Reduce the fraction:
$k^2=\frac{24}{25}.$
Step 7
Choose the positive root and check
Because $k$ is positive, take the positive square root:
$k=\sqrt{\frac{24}{25}}.$
Take the square root of the numerator and denominator:
$k=\frac{\sqrt{24}}{\sqrt{25}}.$
Simplify the denominator:
$k=\frac{\sqrt{24}}{5}.$
Use $24=4\times6$ to simplify the radical:
$k=\frac{2\sqrt6}{5}.$
For a Desmos check, type $k_1^2\sim\frac{24}{25}\{k_1>0\}$ ; the $\sim$ asks Desmos to find a value, and the braces keep it positive. Desmos reports $k_1\approx0.9798$ under PARAMETERS. Then type the exact value for $k$ and both original equations. The line touches the circle once, so the required positive value is $\frac{2\sqrt6}{5}$ . Choice D.

Question 964·Hard·Math

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