Intercepts and Curves

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What is the equation of a circle that has its center at $(5, \frac{1}{4})$ and a radius length of $\frac{1}{5}$ ?

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{10}$

$(x-\frac{1}{4})^2+(y+5)^2=\frac{1}{25}$

$(x-5)^2-(y+\frac{1}{4})^2=\frac{1}{5}$

Explanation

Recall the standard form for the equation of a circle:

In this equation, represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

A straight line passes through the points and .

What is the -intercept of this line?

Explanation

First calculate the slope:

$(\text{change in Y})/(\text{change in x}) = (1-2)/(3-2) = -1$

The standard equation for a line is .

In this equation, is the slope of the line, and is the -intercept. All points on the line must fit this equation. Plug in either point (1,3) or (2,2).

Plugging in (1,3) we get .

Therefore, .

Our equation for the line is now:

To find the -intercept, we plug in :

Thus, the -intercept the point (4,0).

A straight line passes through the points and .

What is the -intercept of this line?

Explanation

First calculate the slope:

$(\text{change in Y})/(\text{change in x}) = (1-2)/(3-2) = -1$

The standard equation for a line is .

In this equation, is the slope of the line, and is the -intercept. All points on the line must fit this equation. Plug in either point (1,3) or (2,2).

Plugging in (1,3) we get .

Therefore, .

Our equation for the line is now:

To find the -intercept, we plug in :

Thus, the -intercept the point (4,0).

Find the y-intercept of the line .

$\left(0, -\frac{6}{5}\right)$

$\left(0, \frac{5}{6}\right)$

$\left(0, \frac{4}{5}\right)$

Explanation

Recall that at the y-intercept, the line crosses the y-axis. This means that at the y-intercept, the value of the x-coordinate is .

Plug in for into the equation to find the y-intercept.

$y=-\frac{6}{5}$

The y-intercept for this line is $\left(0, -\frac{6}{5}\right)$ .

Find the x-intercept of the line .

$(\frac{3}{2}, 0)$

$\left(\frac{3}{4}, 0\right)$

Explanation

Recall that an x-intercept occurs when the line crosses the x-axis. Thus, the y-coordinate of the x-intercept must be .

Plug in for the y-value.

The x-intercept is located at .

What is the y-intercept of the line with the equation ?

Explanation

You should recognize that the given equation is in the point-slope form.

In order to find the y-intercept, rearrange the equation into slope-intercept form, .

Since, , the y-intercept must be located at

What is the y-intercept of a line with the equation ?

Explanation

You should recognize that the given equation is in the point-slope form.

In order to find the y-intercept, rearrange the equation into slope-intercept form, .

Since, , the y-intercept must be located at

Find the equation of a circle if the radius of the circle is and the center is located at the origin.

$x^2+y^2 = \frac{1}{2}$

$x^2+y^2 = \sqrt2$

Explanation

The formula for the equation of a circle is:

The values of $\left ( h,k\right )$ represent the center, and both values are zero at the origin.

Plug in the known values and reduce.

Find the y-intercept of the line .

$\left(0, -\frac{7}{3}\right)$

$\left(0, \frac{3}{7}\right)$

$\left(0, -\frac{5}{3}\right)$

Explanation

Recall that at the y-intercept, the line crosses the y-axis. This means that at the y-intercept, the value of the x-coordinate is .

Plug in for into the equation to find the y-intercept.

$y=-\frac{7}{3}$

The y-intercept for this line is $\left(0, -\frac{7}{3}\right)$ .

Find the x-intercept(s) for the circle $\small (x-4)^2 + y^2 = 9$

$\small 1, 7$

$\small -1, 7$

The circle never intersects the x-axis

$\small -5, 5$

$\small 7$

Explanation

The x-intercepts of any curve are the x-values where the curve is intersecting the x-axis. This happens when y = 0. To figure out these x-values, plug in 0 for y in the original equation and solve for x:

$\small (x-4)^2 + 0^2 = 9$ adding 0 or 0 square doesn't change the value