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SSAT Upper Level Math : Circles

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Example Questions

Example Question #311 : Geometry

A square on the coordinate plane has as its vertices the points $\displaystyle (0,0), (0, 8), (8,0),(8,8)$ . Give the equation of a circle inscribed in the square.

Possible Answers:

$(x-4)^{2}+(y-4)^{2} = 16$ $\displaystyle (x-4)^{2}+(y-4)^{2} = 16$

$(x-8)^{2}+(y-8)^{2} = 32$ $\displaystyle (x-8)^{2}+(y-8)^{2} = 32$

$(x-8)^{2}+(y-8)^{2} = 16$ $\displaystyle (x-8)^{2}+(y-8)^{2} = 16$

$(x-8)^{2}+(y-8)^{2} = 64$ $\displaystyle (x-8)^{2}+(y-8)^{2} = 64$

$(x-4)^{2}+(y-4)^{2} = 32$ $\displaystyle (x-4)^{2}+(y-4)^{2} = 32$

Correct answer:

$(x-4)^{2}+(y-4)^{2} = 16$ $\displaystyle (x-4)^{2}+(y-4)^{2} = 16$

Explanation:

Below is the figure with the circle and square in question:

Circle on axes

The center of the inscribed circle coincides with that of the square, which is the point (4,4) $\displaystyle (4,4)$ . Its diameter is equal to the sidelength of the square, which is 8, so, consequently, its radius is half this, or 4. Therefore, in the standard form of the equation,

$(x-h)^{2}+(y-k)^{2} = r^{2}$ $\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$ ,

substitute $\displaystyle h = k = 4$ and r = 4 $\displaystyle r = 4$ .

$(x-4)^{2}+(y-4)^{2} = 4^{2}$ $\displaystyle (x-4)^{2}+(y-4)^{2} = 4^{2}$

$(x-4)^{2}+(y-4)^{2} = 16$ $\displaystyle (x-4)^{2}+(y-4)^{2} = 16$

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Example Question #252 : Coordinate Geometry

Give the area of the circle on the coordinate plane whose equation is

$x^{2}+ y^{2}+ 4x- 8y -11 = 0$ $\displaystyle x^{2}+ y^{2}+ 4x- 8y -11 = 0$ .

Possible Answers:

$21 \pi$ $\displaystyle 21 \pi$

$20 \pi$ $\displaystyle 20 \pi$

$18 \pi$ $\displaystyle 18 \pi$

$31 \pi$ $\displaystyle 31 \pi$

$11 \pi$ $\displaystyle 11 \pi$

Correct answer:

$31 \pi$ $\displaystyle 31 \pi$

Explanation:

The standard form of the equation of a circle is

$(x-h)^{2}+(y-k)^{2} = r^{2}$ $\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$

where $\displaystyle r$ is the radius of the circle.

We can rewrite the equation we are given, which is in general form, in this standard form as follows:

$x^{2}+ y^{2}+ 4x- 8y -11 = 0$ $\displaystyle x^{2}+ y^{2}+ 4x- 8y -11 = 0$

$x^{2}+ y^{2}+ 4x- 8y =11$ $\displaystyle x^{2}+ y^{2}+ 4x- 8y =11$

$\left (x^{2}+ 4x \right ) +\left ( y^{2}- 8y \right ) =11$ $\displaystyle \left (x^{2}+ 4x \right ) +\left ( y^{2}- 8y \right ) =11$

Complete the squares. Since $\left ( \frac{4}{2} \right )^{2} = 4$ $\displaystyle \left ( \frac{4}{2} \right )^{2} = 4$ and $\left ( \frac{-8}{2} \right )^{2} = 16$ $\displaystyle \left ( \frac{-8}{2} \right )^{2} = 16$ , we do this as follows:

$\left (x^{2}+ 4x + 4 \right ) +\left ( y^{2}- 8y + 16\right ) =11 + 4 + 16$ $\displaystyle \left (x^{2}+ 4x + 4 \right ) +\left ( y^{2}- 8y + 16\right ) =11 + 4 + 16$

$(x-2)^{2}+(y-4)^{2} = 31$ $\displaystyle (x-2)^{2}+(y-4)^{2} = 31$

$r^{2}= 31$ $\displaystyle r^{2}= 31$ , and the area of the circle is

$A = \pi r^{2} = 31 \pi$ $\displaystyle A = \pi r^{2} = 31 \pi$

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Example Question #311 : Geometry

Which of the following is the equation of a circle with center at the origin and area $64 \pi$ $\displaystyle 64 \pi$ ?

Possible Answers:

$x^{2}+ y^{2} = 8$ $\displaystyle x^{2}+ y^{2} = 8$

$x^{2}+ y^{2} = 64$ $\displaystyle x^{2}+ y^{2} = 64$

$x^{2}+ y^{2} = 128$ $\displaystyle x^{2}+ y^{2} = 128$

$x^{2}+ y^{2} = 16$ $\displaystyle x^{2}+ y^{2} = 16$

$x^{2}+ y^{2} = 32$ $\displaystyle x^{2}+ y^{2} = 32$

Correct answer:

$x^{2}+ y^{2} = 64$ $\displaystyle x^{2}+ y^{2} = 64$

Explanation:

The standard form of the equation of a circle is

$(x-h)^{2}+(y-k)^{2} = r^{2}$ $\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$ ,

where the center is (h,k) $\displaystyle (h,k)$ and the radius is $\displaystyle r$ .

The center of the circle is the origin, so $\displaystyle h = k = 0$ , and the equation is

$x^{2}+ y^{2} = r^{2}$ $\displaystyle x^{2}+ y^{2} = r^{2}$

for some $\displaystyle r$ .

The area of the circle is $64 \pi$ $\displaystyle 64 \pi$ , so

$A = \pi r^{2} = 50 \pi$ $\displaystyle A = \pi r^{2} = 50 \pi$

$\frac{ \pi r^{2}}{\pi} = \frac{ 64 \pi}{\pi}$ $\displaystyle \frac{ \pi r^{2}}{\pi} = \frac{ 64 \pi}{\pi}$

$r^{2} = 64$ $\displaystyle r^{2} = 64$

We need go no further; we can substitute to get the equation $x^{2}+ y^{2} = 64$ $\displaystyle x^{2}+ y^{2} = 64$ .

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Example Question #531 : Ssat Upper Level Quantitative (Math)

Which of the following is the equation of a circle with center at the origin and circumference $64 \pi$ $\displaystyle 64 \pi$ ?

Possible Answers:

None of the other responses gives the correct answer.

$x^{2}+ y^{2} = 32$ $\displaystyle x^{2}+ y^{2} = 32$

$x^{2}+ y^{2} = 16$ $\displaystyle x^{2}+ y^{2} = 16$

$x^{2}+ y^{2} = 128$ $\displaystyle x^{2}+ y^{2} = 128$

$x^{2}+ y^{2} = 64$ $\displaystyle x^{2}+ y^{2} = 64$

Correct answer:

None of the other responses gives the correct answer.

Explanation:

The standard form of the equation of a circle is

$(x-h)^{2}+(y-k)^{2} = r^{2}$ $\displaystyle (x-h)^{2}+(y-k)^{2} = r^{2}$ ,

where the center is (h,k) $\displaystyle (h,k)$ and the radius is $\displaystyle r$ .

The center of the circle is the origin, so $\displaystyle h = k = 0$ .

The equation will be

$x^{2}+ y^{2} = r^{2}$ $\displaystyle x^{2}+ y^{2} = r^{2}$

for some $\displaystyle r$ .

The circumference of the circle is $64 \pi$ $\displaystyle 64 \pi$ , so

$C = 2 \pi r = 64 \pi$ $\displaystyle C = 2 \pi r = 64 \pi$

$\frac{2 \pi r }{2 \pi }= \frac{64 \pi}{2 \pi }$ $\displaystyle \frac{2 \pi r }{2 \pi }= \frac{64 \pi}{2 \pi }$

r = 32 $\displaystyle r = 32$

$r^{2}= 32^{2} = 1,024$ $\displaystyle r^{2}= 32^{2} = 1,024$

The equation is $x^{2}+ y^{2} = 1,024$ $\displaystyle x^{2}+ y^{2} = 1,024$ , which is not among the responses.

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SSAT Upper Level Math : Circles

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #311 : Geometry

Example Question #252 : Coordinate Geometry

Example Question #311 : Geometry

Example Question #531 : Ssat Upper Level Quantitative (Math)

All SSAT Upper Level Math Resources

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