Question 108·200 Super-Hard SAT Math Questions·Advanced Math

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The system of equations shown has exactly one real solution.

\begin{cases} x^2+y^2=25 \\ y=mx+6 \end{cases}

Which choice is the value of $m^2$ ?

A line-circle system has exactly one real solution when the line is tangent to the circle: they touch at one point. Substitute the line’s expression for $y$ into the circle equation, then use the quadratic’s discriminant to enforce one real $x$ -value. Desmos can find the slope from that condition, but the final step is to report $m^2$ , not $m$ .

Hints

Hint 1
A solution is a point that satisfies both equations. Since the line already tells you what $y$ equals, replace $y$ in the circle equation with that expression. How many real $x$ -values should the resulting equation have?
Hint 2
For a quadratic $ax^2+bx+c=0$ , the discriminant is $b^2-4ac$ . It must be zero for exactly one real solution, because the plus and minus versions of the quadratic formula then give the same value.
Hint 3
Desmos can find a positive value of $m$ from the discriminant equation. A negative value gives the same square, so you only need one sign. What should you type after finding $m$ to answer the question?

Step-by-step

Substitute and use the discriminant
Step 1
Turn the system into one equation
The line says $y=mx+6$ , so replace $y$ in the circle equation: $x^2+(mx+6)^2=25$ . Each $x$ that works gives one point on the line. So one point shared by the graphs means one real $x$ -value here.
Step 2
Expand the squared expression
Expand the whole square, including its middle term $2(mx)(6)$ : $x^2+m^2x^2+12mx+36=25$ . Dropping the $12mx$ term would change the quadratic and its number of solutions.
Step 3
Move the constant to one side
Subtract $25$ from both sides: $x^2+m^2x^2+12mx+11=0$ . Having zero on one side lets you use the quadratic’s discriminant.
Step 4
Identify the quadratic coefficients
Group the $x^2$ terms: $(1+m^2)x^2+12mx+11=0$ . In the form $ax^2+bx+c=0$ , this gives $a=1+m^2$ , $b=12m$ , and $c=11$ . Since $1+m^2>0$ for real $m$ , it is a quadratic.
Step 5
Set the discriminant to zero
The discriminant is $D=b^2-4ac$ , the quantity under the square root in the quadratic formula. Exactly one real solution means $D=0$ : the two possible roots become the same root. Substitute the coefficients:
$(12m)^2-4(1+m^2)(11)=0.$
Step 6
Find one possible slope in Desmos
Type $(12m_1)^2-4(1+m_1^2)(11)\sim0\{m_1>0\}$ . Desmos uses $m_1$ as the unknown and $\sim$ to fit the equality. Under PARAMETERS, it shows $m_1\approx0.663325$ . The positive restriction selects one slope; the negative slope has the same square.
Step 7
Square the slope to answer the question
Type $m_1^2$ . Desmos displays $0.44$ ; its fraction button shows $\frac{11}{25}$ . That is the requested value of $m^2$ , rather than the value of $m$ . Choice A.

Question 108·200 Super-Hard SAT Math Questions·Advanced Math

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