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SSAT Upper Level Math : Geometry

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Example Questions

Example Question #15 : How To Find The Equation Of A Curve

A horizontal parabola on the coordinate plane includes points (-11, -1) (-1,1) , and (10,2) .

Give its equation.

Possible Answers:

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

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

$x=2y^{2}+5y-8$

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

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

Correct answer:

$x=2y^{2}+5y-8$

Explanation:

The standard form of the equation of a horizontal parabola is

$x=ay^{2}+by+ c$

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a \cdot (-1)^{2}+b \cdot (-1)+ c = -11$

a -b + c = -11

$a \cdot (1)^{2}+b \cdot 1+ c = -1$

a+b + c = -1

$a \cdot (2)^{2}+b \cdot 2+ c = 10$

4a+2b+ c = 10

The three-by-three linear system

a -b + c = -11

a+b + c = -1

4a+2b+ c = 10

can be solved by way of the elimination method.

can be found first, by multiplying the first equation by and add it to the second:

a+b + c = -1

$\underline{-a+b - c = 11}$

=10

$2b \div 2 = 10 \div 2$

b = 5

Substitute 5 for in the last two equations to form a two-by-two linear system:

a+5+ c = -1

a+5+ c - 5= -1 - 5

a + c = -6

$4a+2 \cdot 5+ c = 10$

4a+10+ c = 10

4a+10+ c- 10 = 10 - 10

4a +c = 0

The system

a + c = -6

4a +c = 0

can be solved by way of the substitution method;

4a +c = 0

4a +c -4a = 0 -4a

c = -4a

a + (-4a) = -6

-3a = -6

$-3a\div (-3) = -6 \div (-3)$

a = 2

Substitute 2 for in the top equation:

2 + c = -6

2 + c - 2= -6 - 2

c= -8

The equation is $x=2y^{2}+5y-8$ .

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Example Question #241 : Coordinate Geometry

Circle

Give the equation of the above circle.

Possible Answers:

$(x+2)^{2}+ (y+4)^{2}= 25$

None of the other choices is correct.

$(x-2)^{2}+ (y-4)^{2}= 25$

$(x-2)^{2}+ (y-4)^{2}= 5$

$(x+2)^{2}+ (y+4)^{2}= 5$

Correct answer:

$(x-2)^{2}+ (y-4)^{2}= 25$

Explanation:

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

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

The circle has center (2,4) and radius 5, so substitute:

$(x-2)^{2}+ (y-4)^{2}= 5^{2}$

$(x-2)^{2}+ (y-4)^{2}= 25$

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Example Question #242 : Coordinate Geometry

A circle on the coordinate plane has a diameter whose endpoints are (-3, 7) and (9, 5) . Give its equation.

Possible Answers:

$(x-3)^{2}+ (y-6)^{2}=14$

$(x-3)^{2}+ (y-6)^{2}=148$

$(x+3)^{2}+ (y+6)^{2}=37$

$(x-3)^{2}+ (y-6)^{2}=37$

$(x+3)^{2}+ (y+6)^{2}=148$

Correct answer:

$(x-3)^{2}+ (y-6)^{2}=37$

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:

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

$\left ( \frac{-3+9}{2}, \frac{7+5}{2} \right )$

(3, 6)

Therefore, h = 3 and k = 6

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using (3, 6) and (9, 5) . We will concern ourcelves with finding the square of the radius $r^{2}$ :

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

$r^{2}= \left (9-3\right )^{2} + \left ( 5-6\right )^{2}$

$r^{2}=6^{2} + \left (-1\right )^{2}$

$r^{2}=36+ 1= 37$

Substitute:

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

$(x-3)^{2}+ (y-6)^{2}=37$

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Example Question #243 : Coordinate Geometry

Circle

Give the equation of the above circle.

Possible Answers:

$x^{2}-6x + y ^{2}+4y-3=0$

$x^{2}+6x + y ^{2}-4y-3=0$

$x^{2}-4x + y ^{2}+6y+9=0$

$x^{2}-4x + y ^{2}+6y-3=0$

$x^{2}+4x + y ^{2}-6y+9=0$

Correct answer:

$x^{2}+6x + y ^{2}-4y-3=0$

Explanation:

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

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

The circle has center (-3, 2) and radius 4, so substitute:

$(x-(-3))^{2}+ (y-2)^{2}= 4^{2}$

$(x+3)^{2}+ (y-2)^{2}= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4- 16= 16-16$

$x^{2}+6x + y ^{2}-4y-3=0$

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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:

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

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

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

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

Report an Error

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:

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

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

Report an Error

Example Question #251 : Coordinate Geometry

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

Possible Answers:

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

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

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

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

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

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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SSAT Upper Level Math : Geometry

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #15 : How To Find The Equation Of A Curve

Example Question #241 : Coordinate Geometry

Example Question #242 : Coordinate Geometry

Example Question #243 : Coordinate Geometry

Example Question #3 : Circles

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

Example Question #251 : Coordinate Geometry

Example Question #7 : Circles

Example Question #2 : Circles

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

All SSAT Upper Level Math Resources

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