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Precalculus : Circles

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Example Questions

Example Question #21 : Circles

Write the equation for x^2 +y^2 -6x+8y +4 =0 in standard form

Possible Answers:

(x-3)^2 +(y+4)^2 = 21

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

(x+3)^2 +(y+4)^2 = 29

(x-6)^2 + (y-8)^2 = 4

(x-3)^2 + (y-4)^2 =21

Correct answer:

(x-3)^2 +(y+4)^2 = 21

Explanation:

To determine the standard-form equation, we'll have to complete the square for both x and y. It will be really helpful to re-group our terms to do that:

$x^2 -6x \enspace \enspace \enspace + y^2 +8y \enspace \enspace \enspace = -4$

Adding 9 will complete the square for x, since $(\frac{1}{2}(-6))^2 = (-3)^2 = 9$

Adding 16 will complete the square for y, since $(\frac{1}{2}\cdot8)^2 = 4^2 = 16$

x^2 -6x +9 + y^2 +8y +16 = -4 +9 + 16

Now we just need to simplify. Re-write the left side as two binomials squared, and add the numbers on the right side:

(x-3)^2 + (y+4)^2 = 21

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Example Question #21 : Circles

If the center of the circle is (5,-9) and the radius is 6, what is the equation of the circle in standard form?

Possible Answers:

(x-5)^2 - (y+9)^2 =6

(x-5)^2 + (y+9)^2 =6

(x-5)^2 + (y+9)^2 =36

(x+5)^2 - (y-9)^2 =36

(x+5)^2 + (y-9)^2 =36

Correct answer:

(x-5)^2 + (y+9)^2 =36

Explanation:

Write the equation of the standard form of the circle.

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

where (a,b) is the center of the circle and is the radius.

Substitute the center and the radius into the equation.

(x-5)^2 + (y-(-9))^2 = 6^2

Reduce this equation and leave this in standard form.

(x-5)^2 + (y+9)^2 =36

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Example Question #21 : Circles

Determine the equation of a circle whose center is at (1,1) and radius is .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To solve, simply use the formula for a circle as given below.

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

Thus, our answer is:

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

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Example Question #22 : Circles

Determine the equation for a circle in standard form, centered at (3,-4), with radius 2.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Recall that the standard from for the equation of a circle is

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

where (h, k) is the center, and r is the radius. We are given the center (3, -4) and radius 2. Therefore, h = 3, k = -4, and r = 2. Plugging these vaules into the equation gives us

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

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

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Example Question #25 : Circles

A circle centered at (6,1) passes through (11,13). Write an equation for the circle in standard form.

Possible Answers:

(x+6)^2+(y+1)^2=144

(x-6)^2+(y-1)^2=144

(x-6)^2+(y-1)^2=121

(x+6)^2+(y+1)^2=169

(x-6)^2+(y-1)^2=169

Correct answer:

(x-6)^2+(y-1)^2=169

Explanation:

Recall the equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2 , where (h, k) is the center and r is the radius.

In this problem, we are given the center, but no radius. We must use the other piece of information to find the radius. The second point given is a point on the circle. The definition of a radius is the distance between the center and any point on the circle. Therefore, the radius is equal to the distance between (6,1) and (11,13). Using the distance formula,

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$d=\sqrt{(11-6)^2+(13-1)^2}$

$d=\sqrt{(5)^2+(12)^2}$

$d=\sqrt{25+144}$

d=13

Therefore, the radius is 13. Plugging all the information into the standard form of a circle gives us

(x-6)^2+(y-1)^2=13^2

(x-6)^2+(y-1)^2=169

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Example Question #81 : Conic Sections

Find the equation of a circle whose center is at (0,0) and radius is 3.

Possible Answers:

x^2+y^2-3=0

x^2+y^2-9=0

x^2+y^2=9

x^2+y^2=3

Correct answer:

x^2+y^2=9

Explanation:

To solve, simply plug in the given information into the formula for a circle.

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

Our center is at,

(h,k)=(0,0)

and our radius is 3 therefore,

r^2=3^2=9

Thus,

$x^2+y^2=3^2\Rightarrow x^2+y^2=9$

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Example Question #82 : Conic Sections

Find the equation of a circle with its vertex at the origin and a radius of 8.

Possible Answers:

x^2+y^2-64=0

x^2+y^2=8

x^2+y^2=64

x^2+y^2-8=0

Correct answer:

x^2+y^2=64

Explanation:

To solve, simply remember standard form

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

Also, the center at the origin implies the point (h,k)=(0,0) .

Thus,

x^2+y^2=64

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Example Question #28 : Circles

Write the equation for a circle with a center at (3,4) and a diameter of 7.

Possible Answers:

$(x-3)^2 + (y-4)^2 = \frac{7}{2}$

(x-3)^2 + (y-4)^2 = 49

$(x-3)^2 + (y-4)^2 = \frac{49}{4}$

(x-3)^2 + (y-4)^2 = 7

Correct answer:

$(x-3)^2 + (y-4)^2 = \frac{49}{4}$

Explanation:

The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius.

The diameter is twice the radius:

$r = \frac{d}{2} = \frac{7}{2}$

So plugging in the coordinates the circle equations is:

$(x-3)^2 + (y-4)^2 =(\frac{7}{2})^2 = \frac{49}{4}$

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Example Question #83 : Conic Sections

Write the equation for a circle with a center at (-2,4) and a diameter of 12.

Possible Answers:

(x+2)^2 + (y-4)^2 = 144

(x-2)^2 + (y-4)^2 = 144

(x+2)^2 + (y-4)^2 = 36

(x-2)^2 + (y-4)^2 = 36

Correct answer:

(x+2)^2 + (y-4)^2 = 36

Explanation:

The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius.

The diameter is twice the radius:

$r = \frac{d}{2} = \frac{12}{2} = 6$

So plugging in the coordinates the circle equations is:

(x+2)^2 + (y-4)^2 =6^2 = 36

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Example Question #21 : Circles

Write the equation for a circle with a center at (2,3) and a radius of 4.

Possible Answers:

(x+2)^2 + (y+3)^2 = 16

(x-2)^2 + (y-3)^2 = 16

(x-2)^2 + (y+3)^2 = 16

(x+2)^2 + (y-3)^2 = 16

Correct answer:

(x-2)^2 + (y-3)^2 = 16

Explanation:

The general equation for a circle is:

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

The coordinates of the center of the circle is (h,k) and r is the radius.

So plugging in the coordinates the circle equations is:

(h,k)=(2,3) and r=4

(x-2)^2 + (y-3)^2 =4^2 = 16

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Precalculus : Circles

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Circles

Example Question #21 : Circles

Example Question #21 : Circles

Example Question #22 : Circles

Example Question #25 : Circles

Example Question #81 : Conic Sections

Example Question #82 : Conic Sections

Example Question #28 : Circles

Example Question #83 : Conic Sections

Example Question #21 : Circles

All Precalculus Resources

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