The system of equations shown has exactly one real solution. Which choice is the value of ?

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Super-Hard SAT Math Question 108 | Free SAT Prep | Aniko

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

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The system of equations shown has exactly one real solution. Р‌roper‌tу of Аnikο‌.aі

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

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

When a system is a circle and a line, “exactly one real solution” is your cue for tangency. The fastest algebra method is to set the distance from the circle’s center to the line equal to the radius. Write the line as $Ax+By+C=0$ , use distance $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ , and solve for the requested expression (here, $m^2$ ), which avoids dealing with both possible signs of $m$ . Рower⁠ed bу Аnі⁠kο

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Connect the number of solutions to geometry

A line and a circle can intersect in 0, 1, or 2 points. What does “exactly 1” intersection mean?

Use a tangency condition

For a circle, a line is tangent when the distance from the center to the line equals the radius.

Rewrite the line for the distance formula

Put $y=mx+6$ into the form $Ax+By+C=0$ , then compute the distance from $(0,0)$ . Anik⁠o - F‍ree⁠ ЅАT Prep

Desmos Guide

Graph the circle and the line with a slider

Enter the circle: x^2 + y^2 = 25.

Enter the line: y = m x + 6. When prompted, create a slider for $m$ .

Find when there is exactly one intersection

Move the slider for $m$ until the line just touches the circle (the graphs meet at one point instead of crossing at two points or not meeting at all).

Estimate $m^2$ and match to a choice

When the line is tangent, the slider will show $m$ is about $0.66$ or about $-0.66$ . Anіkο ‌Q⁠uestіon Вank

Square that value (approximately $0.44$ ) and choose the option that equals $0.44$ .

Step-by-step Explanation

Interpret “exactly one real solution”

The equation $x^2+y^2=25$ is a circle centered at $(0,0)$ with radius $5$ .

The system has exactly one real solution when the line $y=mx+6$ touches the circle at exactly one point (is tangent). ani⁠ko.aі/ѕаt

Write the line in standard form and use distance to a line

Rewrite $y=mx+6$ as

mx-y+6=0.

The distance from the center $(0,0)$ to the line $Ax+By+C=0$ is

\frac{|A\cdot 0+B\cdot 0+C|}{\sqrt{A^2+B^2}}.

Here, $A=m$ , $B=-1$ , and $C=6$ , so the distance is

\frac{|6|}{\sqrt{m^2+1}}=\frac{6}{\sqrt{m^2+1}}.

Set the distance equal to the radius and solve for $m^2$

For tangency, the distance from the center to the line must equal the radius $5$ :

\frac{6}{\sqrt{m^2+1}}=5.

Square both sides:

\frac{36}{m^2+1}=25.

36=25(m^2+1)=25m^2+25

and

25m^2=11 \quad\Rightarrow\quad m^2=\frac{11}{25}.

Therefore, the correct choice is $\frac{11}{25}$ .

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Step-by-step Explanation

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