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Precalculus : Circles

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Example Questions

Example Question #1 : Determine The Equation Of A Circle In Standard Form

Which choice would be the circle y^2 + x^2 + 2y - 6x + 1 = 0 in standard form?

Possible Answers:

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

(y+1)^2 + (x-3)^2 = 9

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

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

(y+1)^2 + (x+3)^2 = 9

Correct answer:

(y+1)^2 + (x-3)^2 = 9

Explanation:

To find the standard form of this equation, we have to complete the square for both x and y. It's easiest to do this if we group together the y terms first, then the left terms, and subtract the constant from both sides:

original: y^2 + x^2 + 2y - 6x + 1 = 0 subtract 1 from both sides; group x and y

$y^2 + 2y \enspace \enspace + x^2 -6x \enspace \enspace = -1$

To make the y terms into a square, we have to add 1, since half of 2 is 1, and 1^2 = 1 .

To make the x terms into a square, we have to add 9, since half of -6 is -3, and (-3)^2 = 9 :

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

Now we just have to re-write the y and x terms as the squares that they are, and simplify the right side:

(y+1)^2 + (x-3)^2 = 9

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Example Question #1 : Determine The Equation Of A Circle In Standard Form

Give the center and radius for the circle y^2 - 14y + x^2 - 4 x + 37 = 0 .

Possible Answers:

Center: (-2, -7) ; Radius

Center: (4, 14) ; Radius: $\sqrt{37}$

Center: (4, 7) ; Radius: $\sqrt{28}$

Center: (2,7) ; Radius:

Center: (-7, 2) ; Radius: $\sqrt{8}$

Correct answer:

Center: (2,7) ; Radius:

Explanation:

To determine the center and radius, put this equation into standard form. Standard form is (x-h)^2 + (y-k)^2 = r^2 , where (h,k) is the center and is the radius.

First, group the y terms together and the x terms together, and subtract 37 from both sides:

$y^2 - 14 y \enspace \enspace + x^2 - 4 x \enspace \enspace = -37$

We're leaving a space after these terms because we're completing the square. To make the y terms into a square, we need to add 49, since half of -14 is -7, and (-7)^2 = 49

To make the x terms into a square, we need to add 4, since half of -4 is -2, and (-2)^2 = 4

We'll need to add these to both sides to keep both sides equal:

y^2 - 14 y + 49 + x^2 - 4x + 4 = -37 + 49 + 4

Now we can re-write the quadratics on the left as squares, and simplify the right side:

(y-7)^2 + (x-2)^2 = 16

Interpreting this equation, we can see that the center would be at (2, 7) and the radius is $\sqrt{16} = 4$

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Example Question #9 : Determine The Equation Of A Circle In Standard Form

Write the equation for x^2 + y^2 - 14x + 4 y + 43 = 0 in standard form.

Possible Answers:

(x - 7) ^ 2 + (y - 4 ) ^ 2 = 14

(x- 7)^2 + (y - 2 )^2 = 14

(x - 7 ) ^ 2 + (y + 2 )^2 = 10

(x + 7) ^ 2 + (y + 2 ) ^ 2 = 6

(x - 14) ^ 2 + ( y + 4 ) ^ 2 = 6

Correct answer:

(x - 7 ) ^ 2 + (y + 2 )^2 = 10

Explanation:

To write this in standard form, we will have to complete the square for both x and y. To do this more easily, group the x terms, then the y terms, and then subtract 43 from both sides:

$x^ 2 - 14 x \enspace \enspace \enspace + y^2 + 4 y \enspace \enspace \enspace = -43$

To complete the square for x, we have to add $(\frac{-14}{2})^2 = 49$ to both sides; to complete the square for y, we have to add $(\frac{4}{2})^ 2 = 4$ to both sides:

x^2 -14 x + 49 + y^2 + 4y + 4 = -43 + 49 + 4

Condense the left side and add together the numbers on the right:

(x-7)^2 + (y + 2 )^ 2 = 10

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Example Question #1 : Determine The Equation Of A Circle In Standard Form

Write the standard equation of the circle.

$x^{2}+y^{2}+8x-4y-5=0$

Possible Answers:

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

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

$x^{2}+(y+1)^{2}=4^{2}$

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

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

Correct answer:

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

Explanation:

Group the x and y terms on one side of the equation and the constant on the other.

$(x^{2}+8x)+(y^{2}-4y)=5$

Complete the square by taking half of the middle number for each variable and squaring it. Add the number to the other side of the equation.

$(x^{2}+8x+(\frac{8}{2})^{2})+(y^{2}-4y+(\frac{-4}{2})^{2})=5+(\frac{8}{2})^{2}+(\frac{-4}2{})^{2})$

$(x^{2}+8x+16)+(y^{2}-4y+4)=5+16+4$

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

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Example Question #11 : Circles

Find the equation of the circle if it is centered at (a,b) and has a radius of units.

Possible Answers:

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

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

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

(x+a)^2 + (y+b)^2 = c^2

Correct answer:

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

Explanation:

The equation of a circle centered at $\small (h,\,k) \,$ with radius units in standard form is

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

For the circle ceentered at $\small (a,b)$ with radius $\small c$ units has the equation

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

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Example Question #12 : Circles

Find the equation of the circle if it is centered at $\small (6,4)$ and has a radius of $\small 5$ units.

Possible Answers:

$\small (x-6) + (y-4) = 5$

$\small (x-6)^2 + (y-4)^2 = 25$

$\small (x+6)^2 + (y+4)^2 = 25$

Correct answer:

$\small (x-6)^2 + (y-4)^2 = 25$

Explanation:

The equation of a circle centered at $\small (h,\,k) \,$ with radius units in standard form is

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

For the circle ceentered at $\small \small (6,4)$ with radius $\small 5$ units has the equation

$\small (x-6)^2 + (y-4)^2 = 5^2$

$\small (x-6)^2 + (y-4)^2 = 25$

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Example Question #13 : Circles

(3,6) is a point on a circle whose center is at (-2,-4) . What is the standard equation of this circle?

Possible Answers:

(x-3)^2+(y-6)^2=125

(x+2)^2+(y+4)^2=125

(x+3)^2+(y+6)^2=125

None of the other answers.

(x-2)^2+(y-4)^2=125

Correct answer:

(x+2)^2+(y+4)^2=125

Explanation:

The standard form of the equation of a circle is

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

where the center of the circle resides at the point (h,k) .

Given the center of a circle and a point on the rim of the circle, one can use the distance formula to find the radius.

$d=r=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}=\sqrt{(3+2)^2+(6+4)^2}=\sqrt{125}$

Now plug in the point for the center and radius into the standard equation of a circle:

$(x+2)^2+(y+4)^2=(\sqrt{125})^2\rightarrow \mathbf{(x+2)^2+(y+4)^2=125}$

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Example Question #14 : Circles

Rewrite the following in standard form:

x^2+y^2-6x+12y+20=0

Possible Answers:

(x+3)^2+(y-6)^2=4^2

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

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

(x-6)^2+(y+3)^2=4^2

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

Correct answer:

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

Explanation:

To solve for general form, we want to format the given equation as (x-h)^2+(y-k)^2=r^2

$x^2+y^2-6x+12y+20=0\\ x^2-6x+\qquad+y^2+12y+\qquad=-20\\ (x^2-6x+9)+(y^2+12y+36)=-20+9+36\\ (x-3)^2+(y+6)^2=25$

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

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Example Question #15 : Circles

Rewrite the following equation in standard form:

x^2+y^2+4x+8y-29=0

Possible Answers:

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

(x-2)^2+(y-4)^2=7^2

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

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

(x+2)^2+(y+4)^2=7^2

Correct answer:

(x+2)^2+(y+4)^2=7^2

Explanation:

To rewrite in standard form, we must format the equation as (x-h)^2+(y-k)^2=r^2

$x^2+y^2+4x+8y-29=0\\ x^2+4x+\qquad+y^2+8y+\qquad=29\\ (x^2+4x+4)+(y^2+8y+16)=29+4+16\\ (x+2)^2+(y+4)^2=49\\ (x+2)^2+(y+4)^2=7^2$

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Example Question #16 : Determine The Equation Of A Circle In Standard Form

Rewrite the given equation in standard form:

x^2+y^2+14x-6y+54=0

Possible Answers:

(x-3)^2+(y+7)^2=3^2

(x-7)^2+(y+3)^2=2^2

(x+7)^2+(y-9)^2=8^2

(x+7)^2+(y-3)^2=2^2

(x-7)^2+(y+3)^2=1^2

Correct answer:

(x+7)^2+(y-3)^2=2^2

Explanation:

To rewrite in standard form, we must follow the format (x-h)^2+(y-k)^2=r^2

$x^2+y^2+14x-6y+54=0\\ x^2+14x+\qquad+y^2-6y+\qquad=-54\\ (x^2+14x+49)+(y^2-6y+9)=-54+49+9\\ (x+7)^2+(y-3)^2=4\\ (x+7)^2+(y-3)^2=2^2$

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Precalculus : Circles

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Determine The Equation Of A Circle In Standard Form

Example Question #1 : Determine The Equation Of A Circle In Standard Form

Example Question #9 : Determine The Equation Of A Circle In Standard Form

Example Question #1 : Determine The Equation Of A Circle In Standard Form

Example Question #11 : Circles

Example Question #12 : Circles

Example Question #13 : Circles

Example Question #14 : Circles

Example Question #15 : Circles

Example Question #16 : Determine The Equation Of A Circle In Standard Form

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