Question 206·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 &= 25 \\[1.4em] y &= x + t \end{aligned}

where $t$ is a real parameter. For how many integer values of $t$ does the system have exactly two distinct real solution pairs $(x,y)$ ?

Your answer

A line-circle system has two solution pairs when the line crosses the circle, one when it touches, and none when it misses. Graph the equations in Desmos to see what a crossing represents, then substitute the line into the circle. The discriminant gives the exact range of parameter values. Finally, count the integers including zero; a touching point is not two distinct pairs.

Hints

Hint 1
One equation already gives $y$ in terms of $x$ . Substitution means putting $x+t$ wherever the circle equation uses $y$ . That leaves an equation in one variable, $x$ , whose solutions correspond to the points where the graphs meet.
Hint 2
For a quadratic equation, the discriminant $b^2-4ac$ tells you the number of real roots: positive gives two, zero gives one, and negative gives none. Which sign matches two distinct solution pairs?
Hint 3
After finding the condition on $t^2$ , compare nearby integer squares with its cutoff. Count negative values of $t$ , positive values of $t$ , and $t=0$ separately; both signs can work.

Step-by-step

Substitute and count with the discriminant
Step 1
See what two solutions look like
Type $t=0$ , $x^2+y^2=25$ , and $y=x+t$ on separate Desmos lines. At $t=0$ , Desmos shows two crossings. Each crossing is a pair that satisfies both equations. This is one example, not yet the full range of $t$ .
Step 2
Turn the system into one equation
The line says $y=x+t$ , so use substitution: replace $y$ in the circle equation with $x+t$ : $x^2+(x+t)^2=25$ . The line assigns one $y$ to each $x$ , so two different real solutions for $x$ will give two different solution pairs.
Step 3
Expand the square
Expand the whole square $(x+t)^2$ , including its middle term: $x^2+x^2+2tx+t^2=25$ .
Step 4
Combine the squared terms
Combine the two $x^2$ terms: $2x^2+2tx+t^2=25$ .
Step 5
Put zero on one side
Subtract $25$ from both sides to get zero on one side, the form needed for a root-count test: $2x^2+2tx+t^2-25=0$ .
Step 6
Set up the discriminant
A quadratic has an $x^2$ term as its highest power. Here its coefficients are $a=2$ , $b=2t$ , and $c=t^2-25$ . The discriminant $D=b^2-4ac$ is the expression under the square root in the quadratic formula. Two distinct pairs require $D>0$ : a positive square root gives two different $x$ -values, while $D=0$ gives only one. Substitute the coefficients: $D=(2t)^2-4(2)(t^2-25)$ .
Step 7
Simplify the discriminant
Square $2t$ : $D=4t^2-4(2)(t^2-25)$ . Multiply $4(2)$ : $D=4t^2-8(t^2-25)$ . Distribute the negative factor across both terms: $D=4t^2-8t^2+200$ . Combine like terms: $D=200-4t^2$ .
Step 8
Find the condition for two pairs
Require a positive discriminant: $200-4t^2>0$ . Add $4t^2$ to both sides: $200>4t^2$ . Divide both sides by $4$ : $50>t^2$ . Equality would mean the line touches the circle at only one point.
Step 9
Count the integer values
Since $7^2=49<50$ but $8^2=64>50$ , the integers with $t^2<50$ run from $-7$ through $7$ . That's $7$ negative values, zero, and $7$ positive values. Type $7+1+7$ in Desmos; it prints $15$ . So $15$ integer values of $t$ give exactly two distinct real solution pairs. Grid in 15.

Question 206·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