Circle A has the equation . Circle B has the equation . If circle B passes through the center of circle A, which choice is a possible value of ?

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Super-Hard SAT Math Question 124 | Free SAT Prep | Aniko

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Question 124·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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Circle A has the equation $(x-4)^2+(y+2)^2=9$ . Circle B has the equation $(x-(m+1))^2+(y-(2m-1))^2=40$ . If circle B passes through the center of circle A, which choice is a possible value of $m$ ?

When a circle “passes through” a point, immediately substitute that point into the circle’s equation. Here, the key is recognizing the center of circle A from standard form and then treating circle B’s equation as a distance-squared condition that becomes a solvable quadratic in $m$ . Keep the algebra organized, and use the discriminant to simplify the radical cleanly before comparing to the answer choices.

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Identify what “passes through the center” means

Find the center of circle A, then remember: if a circle passes through a point, that point satisfies the circle’s equation.

Substitute carefully

Plug the center point of circle A into circle B by replacing $x$ and $y$ with the center’s coordinates.

Expect a quadratic in $m$

After substituting, simplify until you get an equation in $m$ only. It should reduce to a quadratic equation.

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Graph the two sides as functions of $m$

Let $x$ represent $m$ . Enter

$y=(4-(x+1))^2+(-2-(2x-1))^2$
$y=40$

Find the intersection x-values

Click the intersection points of the two graphs to read the $x$ -coordinates. These $x$ -values are the possible values of $m$ .

Match to the answer choices

Compare the intersection $x$ -values to the four choices; select the choice that equals one of the intersection values.

Step-by-step Explanation

Find the center of circle A

A circle in the form $(x-h)^2+(y-k)^2=r^2$ has center $(h,k)$ .

For circle A: $(x-4)^2+(y+2)^2=9$ , so its center is $(4,-2)$ .

Use “passes through the center”

If circle B passes through the center of circle A, then the point $(4,-2)$ lies on circle B.

Substitute $x=4$ and $y=-2$ into $(x-(m+1))^2+(y-(2m-1))^2=40$ :

(4-(m+1))^2+(-2-(2m-1))^2=40

Simplify to a quadratic equation

Simplify each term:

$4-(m+1)=3-m$
$-2-(2m-1)=-1-2m$

(3-m)^2+(-1-2m)^2=40

Expand:

(m-3)^2+(2m+1)^2=40

(m^2-6m+9)+(4m^2+4m+1)=40

5m^2-2m+10=40 \\[0.7em] \Rightarrow 5m^2-2m-30=0

Solve and choose a listed solution

Use the quadratic formula on $5m^2-2m-30=0$ :

\begin{aligned} \Delta &= (-2)^2-4(5)(-30)=4+600=604=4\cdot 151 \\ m &= \frac{2\pm \sqrt{604}}{10}=\frac{2\pm 2\sqrt{151}}{10}=\frac{1\pm \sqrt{151}}{5} \end{aligned}

Only one answer choice matches a value of the form $\frac{1\pm\sqrt{151}}{5}$ , so the correct choice is $\frac{1+\sqrt{151}}{5}$ .

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Step-by-step Explanation

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Step-by-step Explanation