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SSAT Upper Level Math : Coordinate Geometry

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Example Questions

Example Question #3 : Circles

A circle on the coordinate plane has a diameter whose endpoints are (2, 7) and (-8, 3) . Give its equation.

Possible Answers:

$x^{2}+6x + y^{2}-10y+5 = 0$

$x^{2}-6x + y^{2}+10y-82= 0$

$x^{2}+6x + y^{2}-10y-82= 0$

$x^{2}-6x + y^{2}+10y+5 = 0$

$x^{2}+6x + y^{2}-10y+63= 0$

Correct answer:

$x^{2}+6x + y^{2}-10y+5 = 0$

Explanation:

A circle with center (h, k) and radius has equation

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

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{2+(-8)}{2}, \frac{7+3}{2} \right )$

(-3, 5)

Therefore, h = -3 and k = 5 .

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using (-3, 5) and (2, 7) . We will concern ourcelves with finding the square of the radius $r^{2}$ :

$r^{2}= \left ( x_{2}-x_{1}\right )^{2} + \left ( y_{2}-y_{1}\right )^{2}$

$r^{2}= \left ( 2- (-3)\right )^{2} + \left ( 7-5\right )^{2}$

$r^{2}= 5^{2} +2^{2}$

$r^{2}=25+ 4 = 29$

Substitute:

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

$(x-(-3))^{2}+ (y-5)^{2}= 29$

$(x+3)^{2}+ (y-5)^{2}= 29$

Expand:

$x^{2}+6x+9+ y^{2}-10y+25= 29$

$x^{2}+6x + y^{2}-10y+34= 29$

$x^{2}+6x + y^{2}-10y+34- 29 = 0$

$x^{2}+6x + y^{2}-10y+5 = 0$

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Example Question #2 : How To Find The Equation Of A Circle

A circle on the coordinate plane has center (4, -6) and circumference $20 \pi$ . Give its equation.

Possible Answers:

$(x+4)^{2}+ (y-6)^{2}=10$

$(x+4)^{2}+ (y-6)^{2}=400$

$(x+4)^{2}+ (y-6)^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=10$

Correct answer:

$(x-4)^{2}+ (y+6)^{2}=100$

Explanation:

A circle with center (h, k) and radius has equation

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

The center is (4, -6) , so h = 4, k=-6 .

To find , use the circumference formula:

$2 \pi r = C$

$2 \pi r = 20 \pi$

r = 10

$r^{2} = 100$

Substitute:

$(x-4)^{2}+ (y-(-6))^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=100$

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Example Question #1 : Circles

A circle on the coordinate plane has center (2, 5) and area $36 \pi$ . Give its equation.

Possible Answers:

$(x+2)^{2}+ (y+5)^{2}= 18$

$(x-2)^{2}+ (y-5)^{2}= 36$

$(x-2)^{2}+ (y-5)^{2}= 6$

$(x-2)^{2}+ (y-5)^{2}=18$

$(x+2)^{2}+ (y+5)^{2}= 6$

Correct answer:

$(x-2)^{2}+ (y-5)^{2}= 36$

Explanation:

A circle with center (h, k) and radius has the equation

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

The center is (2, 5) , so h = 2, k = 5 .

The area is $36 \pi$ , so to find $r^{2}$ , use the area formula:

$\pi r^{2}= A$

$\pi r^{2}= 36 \pi$

$r^{2}= 36$

The equation of the line is therefore:

$(x-2)^{2}+ (y-5)^{2}= 36$

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Example Question #7 : Circles

What is the equation of a circle that has its center at (-4,5) and has a radius of ?

Possible Answers:

(x+4)^2+(y-5)^2=16

(x-4)^2+(y-5)^2=16

(x+4)^2+(y+5)^2=16

(x-4)^2+(y+5)^2=16

Correct answer:

(x+4)^2+(y-5)^2=16

Explanation:

The general equation of a circle with center (a,b) and radius is:

(x-a)^2+(y-b)^2=r^2

Now, plug in the values given by the question:

(x-(-4))^2+(y-5)^2=4^2

(x+4)^2+(y-5)^2=16

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Example Question #2 : Circles

If the center of a circle with a diameter of 5 is located at (2,-3) , what is the equation of the circle?

Possible Answers:

(x-2)^2+(y+3)^2=5

(x+2)^2+(y-3)^2=6.25

(x+2)^2+(y-3)^2=5

(x-2)^2+(y+3)^2=25

(x-2)^2+(y+3)^2=6.25

Correct answer:

(x-2)^2+(y+3)^2=6.25

Explanation:

Write the formula for the equation of a circle with a given point, (h,k) .

(x-h)^2+(y-k)^2=r^2

The radius of the circle is half the diameter, or 2.5 .

Substitute all the values into the formula and simplify.

(x-2)^2+(y-(-3))^2=(2.5)^2

(x-2)^2+(y+3)^2=6.25

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Example Question #9 : How To Find The Equation Of A Circle

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

$x^{2}+ y^{2}+ 4x- 6y + 2 = 0$

Possible Answers:

$2 \pi \sqrt{11}$

$2 \pi \sqrt{2}$

$\pi \sqrt{13}$

$22 \pi$

$2 \pi$

Correct answer:

$2 \pi \sqrt{11}$

Explanation:

The standard form of the equation of a circle is

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

where 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:

$x^{2}+ y^{2}+ 4x- 6y + 2 = 0$

$x^{2}+ y^{2}+ 4x- 6y = -2$

$\left (x^{2}+ 4x \right )+\left ( y^{2}- 6y \right )= -2$

Complete the squares. Since $\left ( \frac{4}{2} \right )^{2} = 4$ and $\left ( \frac{-6}{2} \right )^{2} = 9$ , we do this as follows:

$\left (x^{2}+ 4x +4 \right )+\left ( y^{2}- 6y +9\right )= -2+4+9$

$(x+2)^{2}+(y-3)^{2} = 11$

$r^{2} = 11$ , so $r = \sqrt{11}$ , and the circumference of the circle is

$C = 2 \pi r = 2 \pi \sqrt{11}$

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Example Question #10 : How To Find The Equation Of A Circle

A square on the coordinate plane has as its vertices the points (0,0), (0, 6), (6,0),(6,6) . Give the equation of a circle circumscribed about the square.

Possible Answers:

$(x-3)^{2}+(y-3)^{2} = 36$

$(x-6)^{2}+(y-6)^{2} = 18$

$(x-6)^{2}+(y-6)^{2} = 36$

$(x-3)^{2}+(y-3)^{2} = 18$

$(x-3)^{2}+(y-3)^{2} = 9$

Correct answer:

$(x-3)^{2}+(y-3)^{2} = 18$

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 (3,3) . Its diameter is the length of a diagonal of the square, which is $\sqrt{2}$ times the sidelength 6 of the square - this is $6 \sqrt{2}$ . Its radius is, consequently, half this, or $3 \sqrt{2}$ . Therefore, in the standard form of the equation,

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

substitute h = k = 3 and $r = 3 \sqrt{2}$ .

$(x-3)^{2}+(y-3)^{2} = \left (3 \sqrt{2} \right ) ^{2}$

$(x-3)^{2}+(y-3)^{2} = 3^{2} \left ( \sqrt{2} \right ) ^{2}$

$(x-3)^{2}+(y-3)^{2} = 9 \cdot 2$

$(x-3)^{2}+(y-3)^{2} = 18$

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Example Question #311 : Geometry

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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) . 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,

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

substitute h = k = 4 and r = 4 .

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

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

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Example Question #252 : Coordinate Geometry

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

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

Possible Answers:

$21 \pi$

$20 \pi$

$18 \pi$

$31 \pi$

$11 \pi$

Correct answer:

$31 \pi$

Explanation:

The standard form of the equation of a circle is

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

where 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:

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

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

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

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

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

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

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

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

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Example Question #253 : Coordinate Geometry

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The standard form of the equation of a circle is

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

where the center is (h,k) and the radius is .

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

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

for some .

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

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

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

$r^{2} = 64$

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

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SSAT Upper Level Math : Coordinate Geometry

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #3 : Circles

Example Question #2 : How To Find The Equation Of A Circle

Example Question #1 : Circles

Example Question #7 : Circles

Example Question #2 : Circles

Example Question #9 : How To Find The Equation Of A Circle

Example Question #10 : How To Find The Equation Of A Circle

Example Question #311 : Geometry

Example Question #252 : Coordinate Geometry

Example Question #253 : Coordinate Geometry

All SSAT Upper Level Math Resources

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