For a real number , consider the system of equations: Which choice is the greatest possible value of for which the system has exactly one real solution ?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 112 | Free SAT Prep | Aniko

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Question 112·200 Super-Hard SAT Math Questions·Advanced Math

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For a real number $k$ , consider the system of equations:

\begin{aligned} &x^2+y^2-12x+4y-9=0 \\ &4x+3y=k \end{aligned}

Which choice is the greatest possible value of $k$ for which the system has exactly one real solution $(x,y)$ ?

When a system combines a circle and a line and asks for “exactly one solution,” look for tangency. Rewrite the circle by completing the square to get its center and radius, then use the point-to-line distance formula and set that distance equal to the radius. If two parameter values result, the question will typically ask for the greater or lesser one.

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Identify the first equation

Try rewriting $x^2+y^2-12x+4y-9=0$ by completing the square in both $x$ and $y$ .

Think about “exactly one solution”

A line and a circle intersect at exactly one point when the line is tangent to the circle.

Use distance to a line

Find the distance from the circle’s center to the line $4x+3y=k$ and set it equal to the circle’s radius.

Desmos Guide

Graph the circle

Enter the circle in standard form:

(x-6)^2+(y+2)^2=49

Graph the line with a slider

Enter the line as:

4x+3y=k

Create the slider for $k$ .

Find when there is exactly one intersection

Move the slider for $k$ upward until the line changes from intersecting the circle at two points to not intersecting it at all. The transition occurs at tangency (exactly one intersection). Record the largest $k$ value at that transition.

Step-by-step Explanation

Rewrite the circle in standard form

Complete the square in $x$ and $y$ :

\begin{aligned} &x^2-12x+y^2+4y-9=0 \\ &(x-6)^2-36+(y+2)^2-4-9=0 \\ &(x-6)^2+(y+2)^2=49 \end{aligned}

So the circle has center $(6,-2)$ and radius $7$ .

Use the tangency condition with distance to a line

The line is $4x+3y=k$ . The distance from the center $(6,-2)$ to this line is

\frac{|4(6)+3(-2)-k|}{\sqrt{4^2+3^2}} = \frac{|24-6-k|}{5} = \frac{|18-k|}{5}

For the system to have exactly one real solution, the line must be tangent to the circle, so this distance must equal the radius $7$ .

Solve for the possible values of $k$ and choose the greatest

Set the distance equal to $7$ :

\frac{|18-k|}{5}=7 \quad\Rightarrow\quad |18-k|=35

So $18-k=35$ or $18-k=-35$ , giving $k=-17$ or $k=53$ . The greatest possible value is 53.

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation