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Intermediate Geometry : Coordinate Geometry

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Example Questions

Example Question #26 : How To Find The Equation Of A Circle

Which of the following circles has the same center as the circle with the equation x^2-6x+y^2-10y+28=0 ?

Possible Answers:

(x-3)^2-(y+5)^2=1

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

(x+3)^2-(y+5)^2=4

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

Correct answer:

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

Explanation:

Recall the standard form of the equation of a circle that has a center at (h,k) and a radius of :

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

First, rewrite the given equation of the circle into the standard form of the equation of a circle by completing the squares.

x^2-6x+y^2-10y+28=0

x^2-6x+9+y^2-10y+25+28=9+25

x^2-6x+9+y^2-10y+25=6

(x-3)^2+(y-5)^2=6

The circle then has a center at (3, 5) .

The circle with the equation (x-3)^2+(y-5)^2=2 is the only circle that also has its center at (3, 5) .

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Example Question #27 : How To Find The Equation Of A Circle

What is the equation of a circle that has a center at (6, -1) and a radius of ?

Possible Answers:

(x-6)^2+(y+1)^2=16

(x-6)^2-(y-1)^2=16

(x+6)^2+(y-1)^2=16

(x+6)^2+(y+1)^2=16

Correct answer:

(x-6)^2+(y+1)^2=16

Explanation:

Recall the standard form of the equation of a circle that has a center at (h,k) and a radius of :

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

Plug in the given radius and the center to find the equation of the circle.

(x-6)^2+(y-(-1))^2=4^2

Simplify to reach the standard form of the equation of a circle.

(x-6)^2+(y+1)^2=16

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Example Question #28 : How To Find The Equation Of A Circle

Which of the following circles has the same center as the circle with the equation x^2+2x+y^2-6y-7=0 ?

Possible Answers:

(x-1)^2+(y-3)^2=4

(x+1)^2+(y-3)^2=3

(x+1)^2+(y+3)^2=15

(x-1)^2-(y+3)^2=12

Correct answer:

(x+1)^2+(y-3)^2=3

Explanation:

Recall the standard form of the equation of a circle that has a center at (h,k) and a radius of :

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

First, rewrite the given equation of the circle into the standard form of the equation of a circle by completing the squares.

x^2+2x+y^2-6y-7=0

x^2+2x+1+y^2-6y+9=7+1+9

(x+1)^2+(y-3)^2=17

The circle has a center at (-1, 3) .

The circle with the equation (x+1)^2+(y-3)^2=3 also has its center at (-1, 3) .

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Example Question #29 : How To Find The Equation Of A Circle

Which of the following circles shares a center with a circle given by the equation x^2+8x+y^2+4x+12=0 ?

Possible Answers:

(x+4)^2-(y-2)^2=18

(x+4)^2+(y-2)^2=1

(x-4)^2+(y-2)^2=15

(x+4)^2+(y+2)^2=12

Correct answer:

(x+4)^2+(y+2)^2=12

Explanation:

Recall the standard form of the equation of a circle that has a center at (h,k) and a radius of :

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

First, rewrite the given equation of the circle into the standard form of the equation of a circle by completing the squares.

x^2+8x+y^2+4x+12=0

x^2+8x+16+y^2+4x+6=-12+16+4

(x+4)^2+(y+2)^2=8

The center of the circle is at (-4, -2) .

The circle with the equation (x+4)^2+(y+2)^2=12 also has its center at (-4, -2) .

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Example Question #31 : Circles

A circle has center (7, -4 ) and radius 9. Which of the following is the standard form of the equation of the circle?

Possible Answers:

$(x+ 7) ^{2} + (y- 4) ^{2} = 9$

$(x+ 7) ^{2} + (y- 4) ^{2} = 81$

None of the other choices gives the correct response.

$(x- 7) ^{2} + (y+4) ^{2} = 81$

Correct answer:

$(x- 7) ^{2} + (y+4) ^{2} = 81$

Explanation:

The standard form of the equation of the circle with center at (h, k) and radius is

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

Set h = 7, k = - 4 , r =9 : the correct equation is

$( x-7) ^{2}+ [y-(-4)]^{2} = 9 ^{2}$

$(x-7) ^{2}+ (y+4)^{2} = 81$

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Example Question #32 : Circles

Give the radius of the circle of the equation

$x^{2}+y^{2}+6x-12y+9=0$ .

Possible Answers:

Correct answer:

Explanation:

Rewrite the equation of the circle in the form

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

The radius of the circle will be .

Move the constant to the right by subtracting 9 from both sides; also, reorganize the variable terms at left so as to place - and -terms together:

$x^{2}+y^{2}+6x-12y+9=0$

$x^{2}+y^{2}+6x-12y+9 - 9 =0 - 9$

$x^{2}+6x +y^{2}-12y = -9$

Next, complete the squares. First, we will put blanks after the second and fourth terms (grouping for the sake of readability:

$\left (x^{2}+6x+ \underline{\; \; \; \; \; \; \; } \right )+\left (y^{2}-12y + \underline{\; \; \; \; \; \; \; } \right )= -9$

The blanks will be filled with numbers that will complete two perfect square trinomials. In the first blank will be the square of half of 6, which is

$\left ( \frac{6}{2} \right )^{2} = 3^{2} = 9$ ;

in the second blank, the square of half of , which is

$\left ( \frac{-12}{2} \right )^{2} = 6^{2} = 36$

Add these numbers to both sides:

$(x^{2}+6x+9)+(y^{2}-12y + 36) = -9 + 9 + 36$

$(x^{2}+6x+9)+(y^{2}-12y + 36)= 36$

By construction, the trinomials are perfect squares of binomials, and the expression can be rewritten as:

$(x +3)^{2}+(y-6) ^{2}= 36$

or, since the positive square root of 36 is 6,

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

Therefore, r= 6 , the radius of the circle.

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Example Question #33 : Circles

Circle

Give the equation of the above circle.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The general form of a circle with its center at point (h,k) and with radius is

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

Examine the diagram below:

Circle

The center of the circle is (6,6) and the radius is 6, so set h =k= r = 6 in the circle equation:

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

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

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Example Question #34 : Circles

Circle

Give the equation of the circle in the above diagram.

Possible Answers:

$\left (x+ \frac{9}{2} \right )^{2}+ \left (y+ \frac{9}{2} \right )^{2} = \frac{81}{4}$

$x^{2}+ \left (y- \frac{9}{2} \right )^{2} = \frac{81}{4}$

$x^{2}+ \left (y+ \frac{9}{2} \right )^{2} = \frac{81}{4}$

$\left (x- \frac{9}{2} \right )^{2}+ y^{2} = \frac{81}{4}$

$\left (x+ \frac{9}{2} \right )^{2}+ y^{2} = \frac{81}{4}$

Correct answer:

$x^{2}+ \left (y+ \frac{9}{2} \right )^{2} = \frac{81}{4}$

Explanation:

The general form of a circle with its center at point (h,k) and with radius is

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

The segment with endpoints at the origin and (0, -9) is a diameter of the circle, so the circle has diameter 9, and its radius is half this, or $-\frac{9}{2}$ . The center of the circle is the midpoint of this segment, which is at $\left ( 0, -\frac{9}{2} \right )$ .

The diagram showing the center is below.

Circle

Set $h = 0 , k = -\frac{9}{2}, r = \frac{9}{2}$ . The equation is

$\left [x-0 \right ]^{2}+\left[ y-\left ( - \frac{9}{2} \right ) \right]^{2} = \left ( \frac{9}{2} \right )^{2}$

. $x^{2}+ \left (y+ \frac{9}{2} \right )^{2} = \frac{81}{4}$

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Example Question #35 : Circles

$\textup{Give the coordinates of the center of the circle of the equation}$

$x^{2}+y^{2}+6x-12y+9=0$ .

Possible Answers:

(-3, 6)

(-3, -6)

$\textup{None of the other choices gives the correct response}$

(3, -6)

(3, 6)

Correct answer:

(-3, 6)

Explanation:

Rewrite the equation of the circle in the form

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

The coordinates of the center of the circle will be (h, k) .

Move the constant to the right by subtracting 9 from both sides; also, reorganize the variable terms at left so as to place - and -terms together:

$x^{2}+y^{2}+6x-12y+9=0$

$x^{2}+y^{2}+6x-12y+9 - 9 =0 - 9$

$x^{2}+6x +y^{2}-12y = -9$

Next, complete the squares. First, we will put blanks after the second and fourth terms (grouping for the sake of readability:

$\left (x^{2}+6x+ \underline{\; \; \; \; \; \; \; } \right )+\left (y^{2}-12y + \underline{\; \; \; \; \; \; \; } \right )= -9$

The blanks will be filled with numbers that will complete two perfect square trinomials. In the first blank will be the square of half of 6, which is

$\left ( \frac{6}{2} \right )^{2} = 3^{2} = 9$ ;

in the second blank, the square of half of , which is

$\left ( \frac{-12}{2} \right )^{2} = 6^{2} = 36$

Add these numbers to both sides:

$(x^{2}+6x+9)+(y^{2}-12y + 36) = -9 + 9 + 36$

$(x^{2}+6x+9)+(y^{2}-12y + 36)= 36$

By construction, the trinomials are perfect squares of binomials, and the expression can be rewritten as:

$(x +3)^{2}+(y-6) ^{2}= 36$

or, since the positive square root of 36 is 6,

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

The center of the circle is located at the point (-3, 6) .

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Example Question #36 : Circles

A circle has center (-3, 8 ) and radius . Which of the following is the general form of the equation of the circle?

Possible Answers:

$x^{2}+ y^{2} + 6x - 16y+66 =0$

$\textup{None of the other choices gives the correct response}$

$x^{2}+ y^{2} - 6x + 16y+24 =0$

$x^{2}+ y^{2} - 6x + 16y+66 =0$

$x^{2}+ y^{2}+ 6x - 16y +24 =0$

Correct answer:

$x^{2}+ y^{2}+ 6x - 16y +24 =0$

Explanation:

First, write the standard form of the equation of the circle, which, if it has center at (h, k) and radius , is

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

Set h= -3,k= 8, r =7 - the equation is

$[(x- (-3)]^{2}+( y- 8 )^{2} = 7^{2}$

$(x+3)^{2}+ ( y-8 ) ^{2} = 7^{2}$

The general form of the equation of a circle is

$x^{2}+ y^{2}+ Ax + By+ C = 0$

for some real values of A,B,C .

Expand the squares of the binomials according to the binomial square pattern:

$(x^{2} +2 \cdot x \cdot 3 +3^{2})+(y^{2}- 2 \cdot y \cdot 8 +8 ^{2} )= 7^{2}$

$(x^{2}+ 6x +9 )+(y^{2}-16y +64 )=49$

Rearrange and collect like terms:

$x^{2}+ 6x +9 + y^{2}-16y +64 =49$

$x^{2}+ y^{2} +6x - 16y+9 +64 =49$

$x^{2}+ y^{2}+ 6x - 16y+73 =49$

Subtract 49 from both sides:

$x^{2}+ y^{2}+ 6x - 16y+73 -49 =49 - 49$

$x^{2}+ y^{2}+ 6x - 16y +24 =0$ ,

the general form of the equation.

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Intermediate Geometry : Coordinate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #26 : How To Find The Equation Of A Circle

Example Question #27 : How To Find The Equation Of A Circle

Example Question #28 : How To Find The Equation Of A Circle

Example Question #29 : How To Find The Equation Of A Circle

Example Question #31 : Circles

Example Question #32 : Circles

Example Question #33 : Circles

Example Question #34 : Circles

Example Question #35 : Circles

Example Question #36 : Circles

All Intermediate Geometry Resources

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