Math / Geometry

QuestionWhat is the radius of the circle $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1 ?$ Write your answer in simplified, rationalized form.

Studdy Solution

STEP 1

What is this asking? What's the radius of a circle whose equation is $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1$ ? Watch out! Don't mix up the x-coordinate of the center with the radius!

STEP 2

1. Recall the standard equation of a circle.
2. Identify the radius.

STEP 3

The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle, and $r$ is its radius.
This equation tells us how far each point $(x,y)$ on the circle is from the center!

STEP 4

Our given equation is $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1$ .
Let's rewrite it slightly to match the standard form: $\left(x - \left(-\frac{13}{4}\right)\right)^{2}+(y-0)^{2}=1$ .

STEP 5

Now, we can compare this to the standard equation $(x-h)^2 + (y-k)^2 = r^2$ .
We see that $r^2 = 1$ .
Since the radius must be positive, we take the principal square root of both sides to get $r = \sqrt{1} = 1$ .
So, our radius is 1!

STEP 6

The radius of the circle is 1.

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QuestionWhat is the radius of the circle $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1 ?$ Write your answer in simplified, rationalized form.

Studdy Solution

STEP 1

What is this asking? What's the radius of a circle whose equation is $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1$ ? Watch out! Don't mix up the x-coordinate of the center with the radius!

STEP 2

1. Recall the standard equation of a circle.
2. Identify the radius.

STEP 3

The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle, and $r$ is its radius.
This equation tells us how far each point $(x,y)$ on the circle is from the center!

STEP 4

Our given equation is $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1$ .
Let's rewrite it slightly to match the standard form: $\left(x - \left(-\frac{13}{4}\right)\right)^{2}+(y-0)^{2}=1$ .

STEP 5

Now, we can compare this to the standard equation $(x-h)^2 + (y-k)^2 = r^2$ .
We see that $r^2 = 1$ .
Since the radius must be positive, we take the principal square root of both sides to get $r = \sqrt{1} = 1$ .
So, our radius is 1!

STEP 6

The radius of the circle is 1.

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QuestionWhat is the radius of the circle (x+134)2+y2=1?\left(x+\frac{13}{4}\right)^{2}+y^{2}=1 ?(x+413​)2+y2=1? Write your answer in simplified, rationalized form.

Studdy Solution

STEP 1

What is this asking? What's the radius of a circle whose equation is (x+134)2+y2=1\left(x+\frac{13}{4}\right)^{2}+y^{2}=1(x+413​)2+y2=1? Watch out! Don't mix up the *x*-coordinate of the center with the radius!

STEP 2

1. Recall the standard equation of a circle.2. Identify the radius.

STEP 3

The **standard equation** of a circle is (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2(x−h)2+(y−k)2=r2, where (h,k)(h,k)(h,k) is the **center** of the circle, and rrr is its **radius**.This equation tells us how far each point (x,y)(x,y)(x,y) on the circle is from the center!

STEP 4

Our given equation is (x+134)2+y2=1\left(x+\frac{13}{4}\right)^{2}+y^{2}=1(x+413​)2+y2=1.Let's **rewrite** it slightly to match the standard form: (x−(−134))2+(y−0)2=1\left(x - \left(-\frac{13}{4}\right)\right)^{2}+(y-0)^{2}=1(x−(−413​))2+(y−0)2=1.

STEP 5

Now, we can **compare** this to the standard equation (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2(x−h)2+(y−k)2=r2.We see that r2=1r^2 = 1r2=1.Since the radius must be positive, we take the **principal square root** of both sides to get r=1=1r = \sqrt{1} = 1r=1​=1.So, our **radius** is **1**!

STEP 6

The radius of the circle is **1**.

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Studdy solves anything!

Start learning now

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QuestionWhat is the radius of the circle $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1 ?$ Write your answer in simplified, rationalized form.

What is this asking? What's the radius of a circle whose equation is $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1$ ? Watch out! Don't mix up the x-coordinate of the center with the radius!

1. Recall the standard equation of a circle.
2. Identify the radius.

The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle, and $r$ is its radius.
This equation tells us how far each point $(x,y)$ on the circle is from the center!

Our given equation is $\left(x+\frac{13}{4}\right)^{2}+y^{2}=1$ .
Let's rewrite it slightly to match the standard form: $\left(x - \left(-\frac{13}{4}\right)\right)^{2}+(y-0)^{2}=1$ .

Now, we can compare this to the standard equation $(x-h)^2 + (y-k)^2 = r^2$ .
We see that $r^2 = 1$ .
Since the radius must be positive, we take the principal square root of both sides to get $r = \sqrt{1} = 1$ .
So, our radius is 1!

The radius of the circle is 1.