In the -plane, a circle is tangent to both the -axis and the -axis, and its center is in the first quadrant. The circle passes through the point . An equation of the circle is , where , , and are constants. If the radius of the circle is less than 10, which choice is the value of ?

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Super-Hard SAT Math Question 147 | Free SAT Prep | Aniko

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Question 147·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the $xy$ -plane, a circle is tangent to both the $x$ -axis and the $y$ -axis, and its center is in the first quadrant. The circle passes through the point $(8,2)$ . An equation of the circle is $x^2 + y^2 + ax + by + c = 0$ , where $a$ , $b$ , and $c$ are constants.

If the radius of the circle is less than 10, which choice is the value of $c$ ?

When you see “tangent to the $x$ -axis” or “tangent to the $y$ -axis,” immediately translate tangency into a distance-from-center fact: the radius equals the center’s perpendicular distance to the axis. Here, tangency to both axes with the center in the first quadrant forces the center to be $(r,r)$ , which makes the circle easy to write as $(x-r)^2+(y-r)^2=r^2$ . Use the given point to form a quadratic in $r$ , pick the root that matches any extra condition (like $r<10$ ), and then relate the expanded form to $x^2+y^2+ax+by+c=0$ (in particular, $c=h^2+k^2-r^2$ , which becomes $c=r^2$ in this setup).

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Locate the center from tangency

A circle tangent to an axis has its radius equal to the center’s distance to that axis. If it’s tangent to both axes in the first quadrant, what must the center look like?

Use a single variable

Let the radius be $r$ . Write the center in terms of $r$ , then write the circle as $(x-r)^2+(y-r)^2=r^2$ .

Plug in the given point

Substitute $(8,2)$ into the circle equation to get an equation involving only $r$ . It should simplify to a quadratic.

Connect back to c

After you find the correct value of $r$ (using $r<10$ ), remember that in $x^2+y^2+ax+by+c=0$ , the constant term comes from the expanded form of the circle.

Desmos Guide

Model the radius with a single variable

Let $x$ represent the radius $r$ . Enter

f(x)=(8-x)^2+(2-x)^2-x^2.

This expression equals 0 exactly when $(8,2)$ lies on the circle centered at $(x,x)$ with radius $x$ .

Find the possible radii

Graph y=f(x) and find the $x$ -intercepts. There should be two positive intercepts (two possible radii).

Apply the radius condition and compute c

Choose the intercept with $x<10$ . Then compute $c=x^2$ (you can type c=x^2 or just square the chosen value) and match it to the answer choices.

Step-by-step Explanation

Use the tangency information

If a circle is tangent to both the $x$ -axis and the $y$ -axis and its center is in the first quadrant, then the center must be the same distance from both axes.

Let the radius be $r$ . Then the center is $(r,r)$ , and the circle can be written as

(x-r)^2+(y-r)^2=r^2.

Expanding gives

x^2+y^2-2rx-2ry+r^2=0,

so in the form $x^2+y^2+ax+by+c=0$ , we have $c=r^2$ .

Use the point on the circle to find r

Substitute $(8,2)$ into $(x-r)^2+(y-r)^2=r^2$ :

(8-r)^2+(2-r)^2=r^2.

Expand and simplify:

\begin{aligned} (64-16r+r^2)+(4-4r+r^2) &= r^2 \\ 68-20r+2r^2 &= r^2 \\ r^2-20r+68 &= 0. \end{aligned}

Solve the quadratic and choose the correct radius

Solve $r^2-20r+68=0$ :

\begin{aligned} r &= \frac{20\pm\sqrt{20^2-4\cdot 1\cdot 68}}{2} \\ &= \frac{20\pm\sqrt{400-272}}{2} \\ &= \frac{20\pm\sqrt{128}}{2} \\ &= \frac{20\pm 8\sqrt{2}}{2} \\ &= 10\pm 4\sqrt{2}. \end{aligned}

Because the radius is less than 10, choose $r=10-4\sqrt{2}$ .

Compute c

Since $c=r^2$ ,

c=(10-4\sqrt{2})^2=100-80\sqrt{2}+32=132-80\sqrt{2}.

So the correct choice is $132-80\sqrt{2}$ .

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