Question 955·Hard·Math

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\begin{cases} x^2 + y^2 = 4k \\ y = x + k \end{cases}

In the system of equations above, $k$ is a positive constant. The system has exactly one real solution $(x, y)$ . What is the value of $k$ ?

Your answer

For a line-circle system, exactly one solution means the line touches the circle once instead of crossing it twice. Substitute the line’s expression for $y$ into the circle equation to get a quadratic in $x$ . Its discriminant, the part under the quadratic formula’s square root, must be zero for one root. Then check the stated condition on the constant; an excluded value may also make the discriminant zero.

Hints

Hint 1
A solution to a system is a point on both graphs. If a line crosses a circle, it gives two solutions; if it touches the circle at only one point, it gives one. Which equation already tells you what $y$ equals?
Hint 2
Use substitution: replace $y$ in the circle equation with the expression the line gives for $y$ . You’ll have an equation in $x$ whose number of real roots matches the number of points the graphs share.
Hint 3
For a quadratic $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ is zero when its two roots become one. After setting it to zero, which possible value must you reject because $k$ is positive?

Step-by-step

Substitute and use the discriminant
Step 1
See what one solution would look like
Type $k=1$ , $x^2+y^2=4k$ , and $y=x+k$ on separate Desmos lines. At this sample value, the line crosses the circle at two points. Each shared point is a solution, so exactly one solution would mean a one-point touch. The $k=1$ slider is only a picture, not a value assumed in the algebra.
Step 2
Make one equation in x
The line says $y=x+k$ , so replace $y$ in the circle equation with the whole expression $x+k$ :
$x^2+(x+k)^2=4k.$
Each real $x$ that works gives one point on the line and circle.
Step 3
Expand the square
Expand $(x+k)^2$ , keeping its middle term $2kx$ :
$x^2+x^2+2kx+k^2=4k.$
Step 4
Combine the x-squared terms
Combine the two $x^2$ terms:
$2x^2+2kx+k^2=4k.$
Step 5
Move all terms to one side
Subtract $4k$ from both sides:
$2x^2+2kx+k^2-4k=0.$
Step 6
Put the quadratic in standard form
Divide every term by $2$ :
$x^2+kx+\frac{k^2}{2}-2k=0.$
Step 7
Set the discriminant to zero
In $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . When it is zero, the plus and minus versions of the quadratic formula give the same root, so there is exactly one $x$ -value. Here $a=1$ , $b=k$ , and $c=\frac{k^2}{2}-2k$ , so:
$k^2-4(1)\left(\frac{k^2}{2}-2k\right)=0.$
Step 8
Simplify the condition on k
Distribute the $-4$ through both terms in parentheses:
$k^2-2k^2+8k=0.$
Combine the $k^2$ terms:
$-k^2+8k=0.$
Step 9
Factor to expose both possibilities
Factor out $k$ :
$k(8-k)=0.$
A product is zero when at least one factor is zero. Don’t discard a possibility until you check the condition that $k$ is positive.
Step 10
Solve for the allowed value
Because $k$ is positive, it cannot be the zero factor in $k(8-k)=0$ . Type $8-k_1\sim0$ in Desmos to solve using the other factor. Use $k_1$ as the unknown so it isn’t confused with the $k=1$ picture; $\sim$ asks Desmos to fit the equality. Under PARAMETERS, Desmos shows $k_1=8$ . The given condition excludes $k=0$ . So the positive constant that gives exactly one real solution is $8$ . Set the discriminant to zero for one root, then check which values the problem allows. Grid in 8.

Question 955·Hard·Math

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