Question 171·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Consider the system of equations

\begin{aligned} x^2 + y^2 &= 10 \\[1.4em] y &= x + k \end{aligned}

where $k$ is a positive real constant. Which value of $k$ makes the system have exactly one real solution $(x, y)$ ?

A line and a circle can meet twice, touch once, or miss. Exactly one solution signals tangency: the line touches the circle at one point. Graph the system in Desmos to see what changing the constant does, then substitute the line into the circle. Set the resulting quadratic’s discriminant to zero to find the exact value. A line that crosses the circle has two solutions, even if you spot only one.

Hints

Hint 1
A solution to a system is a point on both graphs. A line that passes through a circle meets it on the way in and on the way out. What must the line do to give exactly one shared point?
Hint 2
The line gives you $y=x+k$ , so replace every $y$ in the circle equation with $(x+k)$ . Each real $x$ in the resulting equation gives one point, because the line determines its $y$ -coordinate.
Hint 3
The discriminant of $ax^2+bx+c=0$ is $b^2-4ac$ . It equals zero when the quadratic has one real root. After you expand the substituted equation, what are $a$ , $b$ , and $c$ ?

Step-by-step

Approach 1: Substitute and count roots
Step 1
See what two solutions look like
Type the circle and line into Desmos, and try $k=2$ from the choices. Click where they meet: Desmos shows $(-3,-1)$ and $(1,3)$ . Each intersection, or shared point, solves both equations. This trial line crosses the circle; you need a value that makes it touch at only one point.
Step 2
Turn the system into one equation
At any shared point, the line says $y=x+k$ . Use substitution: put that whole expression in place of $y$ in the circle equation:
$x^2+(x+k)^2=10.$
Step 3
Write the quadratic in standard form
Expand $(x+k)^2$ :
$x^2+x^2+2kx+k^2=10.$
Combine the $x^2$ terms:
$2x^2+2kx+k^2=10.$
Subtract $10$ so the equation has zero on one side:
$2x^2+2kx+k^2-10=0.$
Step 4
Find the condition for one root
For $ax^2+bx+c=0$ , the discriminant is $D=b^2-4ac$ . Exactly one real root means $D=0$ . Here $a=2$ , $b=2k$ , and $c=k^2-10$ . Each root gives one point on the line, so this tests the system’s solution count. Substitute:
$D=(2k)^2-4(2)(k^2-10).$
Square $2k$ :
$D=4k^2-4(2)(k^2-10).$
Multiply $4(2)$ :
$D=4k^2-8(k^2-10).$
Distribute $-8$ :
$D=4k^2-8k^2+80.$
Combine the $k^2$ terms:
$D=80-4k^2.$
Step 5
Find the positive value of the constant
Set the discriminant to zero:
$80-4k^2=0.$
Add $4k^2$ :
$80=4k^2.$
Divide by $4$ :
$20=k^2.$
Type $k_1^2\sim20\{k_1>0\}$ in Desmos. The $\sim$ makes this a regression, and the restriction keeps the positive solution the question requires. Desmos reports $k_1\approx4.472$ under PARAMETERS. Take the positive square root:
$k=\sqrt{20}.$
Take the perfect-square factor $4$ out of the radical:
$k=2\sqrt5.$
So the line touches the circle at exactly one point when $k=2\sqrt5$ . Choice C.

Approach 2: Use the distance to a tangent line
Step 1
Read the circle’s radius
In $x^2+y^2=r^2$ , the center is $(0,0)$ and $r$ is the radius. Here $r=\sqrt{10}$ . A tangent touches the circle once, so its perpendicular distance from the center must equal $\sqrt{10}$ .
Step 2
Put the line in distance-formula form
Subtract $y$ from both sides of $y=x+k$ :
$x-y+k=0.$
The coefficients of $x$ and $y$ are now $1$ and $-1$ , which lets you find the line’s distance from the origin.
Step 3
Find the center-to-line distance
For a line $ax+by+c=0$ , its perpendicular distance from $(0,0)$ is $\frac{|c|}{\sqrt{a^2+b^2}}$ . Use $a=1$ , $b=-1$ , and $c=k$ :
$d=\frac{|k|}{\sqrt{1^2+(-1)^2}}.$
Since the question says $k$ is positive, simplify:
$d=\frac{k}{\sqrt2}.$
Step 4
Make that distance the radius
For one point of contact, set the distance equal to the radius:
$\frac{k}{\sqrt2}=\sqrt{10}.$
Multiply by $\sqrt2$ :
$k=\sqrt{10}\sqrt2.$
Multiply inside the radical:
$k=\sqrt{20}.$
Take out the perfect-square factor $4$ :
$k=2\sqrt5.$
That positive value makes the line tangent to the circle. Choice C.

Question 171·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

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Step-by-step

Lessons that teach this