A circle in the -plane has radius 5 and passes through the points and . The center of the circle lies below the -axis. An equation of the circle is where , , and are constants. Which choice is the value of ?

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Super-Hard SAT Math Question 197 | Free SAT Prep | Aniko

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Question 197·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A circle in the $xy$ -plane has radius 5 and passes through the points $(1,2)$ and $(5,2)$ . The center of the circle lies below the $x$ -axis. An equation of the circle is

x^2+y^2+ax+by+c=0,

where $a$ , $b$ , and $c$ are constants. Which choice is the value of $c$ ? а⁠nikо.a‍i⁠/⁠ѕat

When two points on a circle share the same $y$ -value, immediately use geometry: the center lies on the perpendicular bisector of the chord, which often pins down one coordinate of the center and therefore one of the coefficients ( $a$ or $b$ ). Then use a known point to relate the remaining coefficients, and finish with the circle identity $r^2=\frac{a^2+b^2}{4}-c$ . If a condition describes where the center is located (like “below the $x$ -axis”), use it to choose the correct root after solving a quadratic. Prοрertу⁠ οf А‌n⁠iкo.ai

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Look for symmetry in the given points

The points $(1,2)$ and $(5,2)$ have the same $y$ -coordinate. What vertical line is the perpendicular bisector of the segment between them?

Connect the center to $a$ and $b$

For $x^2+y^2+ax+by+c=0$ , the center is $\left(-\frac a2,-\frac b2\right)$ . Use the line you found to determine $a$ .

Use one point, then use the radius formula

Plug one given point into the circle equation to express $c$ in terms of $b$ . Then use $r^2=\frac{a^2+b^2}{4}-c$ with $r=5$ .

Use “center below the $x$ -axis” to choose the right root

Once you get two possible values of $b$ , translate “below the $x$ -axis” into an inequality involving $-\frac b2$ . Aniкο.аi‌ - ЅA‌Т Рreр

Desmos Guide

Encode the radius condition as a function of $b$

In Desmos, define a function of $x$ (use $x$ to stand for $b$ ):

f(x)=(36+x^2)/4-(1-2x)

This is the expression for $r^2$ after using $a=-6$ and $c=1-2b$ .

Find $b$ values that make the radius 5

Graph y=f(x) and also graph y=25. C‌оntеnt bу Aniкo.ai

Click the intersection points; their $x$ -coordinates are the possible values of $b$ .

Choose the $b$ that puts the center below the $x$ -axis

The center’s $y$ -coordinate is $-b/2$ , so “below the $x$ -axis” means $b>0$ .

From the two intersection $x$ -values, choose the positive one.

Compute $c$ from $c=1-2b$

With the chosen value of $b$ , compute c=1-2b (you can type 1-2*(that b value) in an expression line).

Match that value to the answer choices.

Step-by-step Explanation

Use symmetry to find $a$

The points $(1,2)$ and $(5,2)$ have the same $y$ -coordinate, so the segment connecting them is horizontal.

The center of a circle lies on the perpendicular bisector of any chord, so the center must lie on the vertical line through the midpoint of these two points.

The midpoint is

\left(\frac{1+5}{2},\frac{2+2}{2}\right)=(3,2),

so the perpendicular bisector is $x=3$ . Therefore the center has $x$ -coordinate $3$ . Р‍o‌werеd by Аnіко

For $x^2+y^2+ax+by+c=0$ , the center is $\left(-\frac a2,-\frac b2\right)$ , so

-\frac a2=3 \quad\Rightarrow\quad a=-6.

Relate $c$ and $b$ using a point on the circle

Substitute $a=-6$ into the equation:

x^2+y^2-6x+by+c=0.

Since $(1,2)$ lies on the circle:

1^2+2^2-6(1)+b(2)+c=0

1+4-6+2b+c=0 \quad\Rightarrow\quad c=1-2b.

Use the radius condition to solve for $b$

For $x^2+y^2+ax+by+c=0$ , the radius satisfies

r^2=\frac{a^2+b^2}{4}-c.

Here $r=5$ , $a=-6$ , and $c=1-2b$ , so

25=\frac{36+b^2}{4}-(1-2b).

Multiply by 4:

100=36+b^2-4+8b

b^2+8b-68=0.

Solve:

\begin{aligned} b&=\frac{-8\pm\sqrt{8^2-4(1)(-68)}}{2} \\ &=\frac{-8\pm\sqrt{64+272}}{2} \\ &=\frac{-8\pm\sqrt{336}}{2} \\ &=\frac{-8\pm 4\sqrt{21}}{2} \\ &=-4\pm 2\sqrt{21}. \end{aligned}

The center is $\left(3,-\frac b2\right)$ . “Below the $x$ -axis” means $-\frac b2<0$ , so $b>0$ .

Thus choose the positive solution $b=-4+2\sqrt{21}$ . (Do not compute $c$ yet.)

Compute $c$

From Step 2, $c=1-2b$ . Substitute $b=-4+2\sqrt{21}$ :

\begin{aligned} c&=1-2(-4+2\sqrt{21}) \\ &=1+8-4\sqrt{21} \\ &=9-4\sqrt{21}. \end{aligned}

So the value of $c$ is $9-4\sqrt{21}$ .

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Step-by-step Explanation

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