In the -plane, a line and a circle are given by the equations If is an intersection point of the line and the circle in the first quadrant, which choice is the value of ?

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Super-Hard SAT Math Question 103 | Free SAT Prep | Aniko

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Question 103·200 Super-Hard SAT Math Questions·Advanced Math

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In the $xy$ -plane, a line and a circle are given by the equations

\begin{aligned} \text{Line: } & y = 2x + 1 \\ \text{Circle: } & x^2 + y^2 = 25 \end{aligned}

If $(x,y)$ is an intersection point of the line and the circle in the first quadrant, which choice is the value of $x$ ?

For nonlinear systems like a line with a circle, substitute the linear equation into the nonlinear one to reduce the system to a single-variable quadratic. Then solve efficiently (often by the quadratic formula) and use the context—in this case, “first quadrant”—to select the correct solution and discard the other root.

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Use both equations at once

An intersection point satisfies both equations. Replace $y$ in the circle equation using $y=2x+1$ .

Make it a quadratic in $x$

After substitution, expand $(2x+1)^2$ and combine like terms to get an equation in the form $ax^2+bx+c=0$ .

Use the first quadrant condition

You should get two possible $x$ -values from the quadratic. Decide which one is valid by using that a first-quadrant point has $x>0$ and $y>0$ .

Desmos Guide

Graph the two equations

Enter the line as y=2x+1 and the circle as x^2+y^2=25.

Find the intersections

Click on an intersection point (or use the intersection points that appear) to see the coordinates of the intersection points.

Select the first-quadrant intersection and read off $x$

Choose the intersection with both coordinates positive. The $x$ -coordinate of that point is $\frac{2(\sqrt{31}-1)}{5}$ .

Step-by-step Explanation

Substitute to create a single equation in one variable

Because the point lies on both graphs, substitute $y=2x+1$ into $x^2+y^2=25$ :

\begin{aligned} x^2+(2x+1)^2&=25 \\ \end{aligned}

Simplify to a quadratic equation

Expand and combine like terms:

\begin{aligned} (2x+1)^2 &= 4x^2+4x+1 \\ x^2+4x^2+4x+1&=25 \\ 5x^2+4x-24&=0 \end{aligned}

Solve the quadratic and choose the first-quadrant solution

Use the quadratic formula with $a=5$ , $b=4$ , and $c=-24$ :

\begin{aligned} x&=\frac{-4\pm\sqrt{4^2-4(5)(-24)}}{2(5)} \\ &=\frac{-4\pm\sqrt{16+480}}{10} \\ &=\frac{-4\pm\sqrt{496}}{10} \\ &=\frac{-4\pm 4\sqrt{31}}{10} \\ &=\frac{-2\pm 2\sqrt{31}}{5} \end{aligned}

One solution is negative, but a point in the first quadrant must have $x>0$ , so

x=\frac{-2+2\sqrt{31}}{5}=\frac{2(\sqrt{31}-1)}{5}.

Therefore, the correct choice is $\frac{2(\sqrt{31}-1)}{5}$ .

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Step-by-step Explanation

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