# Conic Sections And Intro To Circles

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| By Jhoane Perez
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Jhoane Perez
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Quizzes Created: 1 | Total Attempts: 41
Questions: 10 | Attempts: 41

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• 1.

### Which of the following is the center and the radius of this equation x2 +y2 -6x -26y +162 =0

• A.

• B.

• C.

• D.

Explanation
The equation of a circle can be written in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius. In this case, the equation x^2 + y^2 - 6x - 26y + 162 = 0 can be rearranged to (x - 3)^2 + (y - 13)^2 = 4^2. Therefore, the center of the circle is (3, 13) and the radius is 4.

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• 2.

• A.

Option 1

• B.

Option 2

• C.

Option 3

• D.

Option 4

D. Option 4
• 3.

### What is the equation of the circle given that the center is (3,-2) and radius =4

• A.

Option 1

• B.

Option 2

• C.

Option 3

• D.

Option 4

A. Option 1
Explanation
The equation of a circle with center (h, k) and radius r is given by (x-h)^2 + (y-k)^2 = r^2. In this case, the center is (3, -2) and the radius is 4. Plugging these values into the equation, we get (x-3)^2 + (y+2)^2 = 16. Therefore, Option 1 is the correct answer.

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• 4.

### What is the standard form of the equation x2 + y2 + 14x - 12y + 4 =0

• A.

Option 1

• B.

Option 2

• C.

Option 3

• D.

Option 4

A. Option 1
Explanation
The standard form of a quadratic equation is given by Ax^2 + By^2 + Cx + Dy + E = 0, where A, B, C, D, and E are constants. In the given equation, we can rewrite it as x^2 + y^2 + 14x - 12y + 4 = 0. By rearranging the terms, we can see that the equation is in the standard form with A = 1, B = 1, C = 14, D = -12, and E = 4. Therefore, the correct answer is Option 1.

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• 5.

### Given the center at (0,0) and the radius is 5, find the equation of the circle.

• A.

X2 - y2 = 10

• B.

X2 + y2 = 10

• C.

X2 + y2 = 25

• D.

X2 - y2 = 25

C. X2 + y2 = 25
Explanation
The equation of a circle with center (0,0) and radius 5 is x^2 + y^2 = 25. This equation represents all the points (x,y) that are 5 units away from the origin (0,0), forming a circle with a radius of 5.

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• 6.

### This is a curve formed by the intersection of a plane and a double right circular cone.

• A.

Circle

• B.

Conic

• C.

Parabola

• D.

Generator of the cone

B. Conic
Explanation
A conic is the correct answer because it is a curve formed by the intersection of a plane and a double right circular cone. A conic can take different forms depending on the angle and position of the intersecting plane, such as a circle, ellipse, parabola, or hyperbola. In this case, the intersecting plane forms a conic shape, but without further information, we cannot specify which specific type of conic it is.

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• 7.

### This is a set of all coplanar points such that the distance from a fixed point is constant.

• A.

Circle

• B.

Parabola

• C.

Hyperbola

• D.

Ellipse

A. Circle
Explanation
A circle is defined as a set of all coplanar points that are equidistant from a fixed point called the center. This means that the distance from any point on the circle to the center is constant. Therefore, a circle fits the given description of a set of coplanar points with a constant distance from a fixed point.

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• 8.

### What is the equation of the circle whose center is (-6, -15) and radius is square root of 5

• A.

Option 1

• B.

Option 2

• C.

Option 3

• D.

Option 4

B. Option 2
Explanation
The equation of a circle with center (h, k) and radius r is given by (x-h)^2 + (y-k)^2 = r^2. In this case, the center is (-6, -15) and the radius is square root of 5. Therefore, the equation of the circle is (x+6)^2 + (y+15)^2 = 5. Option 2 is the correct answer.

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• 9.

• A.

Option 1

• B.

Option 2

• C.

Option 3

• D.

Option 4

A. Option 1
• 10.

• A.

• B.

• C.

• D.