Circles, Ellipses, and Hyperbolas - SAT Subject Test in Math II

Card 0 of 5

Question

Refer to the above figure. The circle has its center at the origin. What is the equation of the circle?

Answer

The equation of a circle with center and radius is

$(x-h)^{2}+ \left ( y - k\right )^{2} = r^{2}$

The center is at the origin, or , so . To find $r^{2}$ , use the distance formula as follows:

$r^{2} = \left ( 6-0\right )^{2} +\left ( 8 -0 \right )^{2} = 6^{2}+8^{2} = 36 + 64 = 100$

Note that we do not actually need to find .

We can now write the equation of the circle:

$(x-0)^{2}+ \left ( y - 0\right )^{2} = 100$

$x^{2}+y^{2} = 100$

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Question

Refer to the above diagram. The circle has its center at the origin; is the point . What is the length of the arc $\widehat{AB}$ , to the nearest tenth?

Answer

First, it is necessary to determine the radius of the circle. This is the distance between and , so we apply the distance formula:

$r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}$

$r = \sqrt{\left ( -9-0\right )^{2} +\left ( -12 -0 \right )^{2} }$

$r=\sqrt{ (-9)^{2}+(-12)^{2}}$

$r= \sqrt{81+144 }= \sqrt{225} = 15$

The circumference of the circle is

$C = 2 \pi r = 2 \pi \cdot 15 = 30 \pi$

Now we need to find the degree measure of the arc. We can do this best by examining this diagram:

The degree measure of $\widehat{AB}$ is also the measure $\theta$ of the central angle formed by the green radii. This is found using the relationship

$\tan \theta = \frac{y}{x}$

$\tan \theta = \frac{-12}{-9} \approx 1.333$

Using a calculator, we find that $\tan ^{-1} 1.3333 \approx 53.12^{\circ }$ . We can adjust for the location of :

$\theta \approx \left (180 - 53.12 \right )^{\circ } \approx 126.87^{\circ }$

which is the degree measure of the arc.

Now we can calculate the length of the arc:

$L \approx \frac{126.87}{360} \cdot C$

$L \approx \frac{126.87}{360} \cdot 30 \pi \approx 33.2$

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Question

A triangle has its vertices at the points with coordinates , , and . Give the equation of the circle that circumscribes it.

Answer

The circumscribed circle of a triangle is the circle which passes through all three vertices of the triangle.

In general form, the equation of a circle is

$x^{2}+ y^{2}+ Ax + By + C = 0$ .

Since the circle passes through the origin, substitute ; the equation becomes

$0^{2}+0^{2}+ A \cdot 0 + B \cdot 0 + C = 0$

Therefore, we know the equation of any circle passing through the origin takes the form

$x^{2}+ y^{2}+ Ax + By = 0$

for some .

Since the circle passes through , substitute ; the equation becomes

$7^{2}+ 0^{2}+ A \cdot 7 + B \cdot 0 = 0$

Solving for :

Now we know that the equation takes the form

$x^{2}+ y^{2}-7x + By = 0$

for some .

Since the circle passes through , substitute ; the equation becomes

$4^{2}+ 4^{2}- 7 \cdot 4 + B \cdot 4 = 0$

Solving for :

The general form of the equation of the circle is therefore

$x^{2}+ y^{2}-7x -y = 0$

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Question

On the coordinate plane, the vertices of a square are at the points with coordinates . Give the equation of the circle.

Answer

The figure in question is below.

The center of the circle can be seen to be the origin, so, if the radius is , the equation will be

$x^{2}+ y^{2} = r^{2}$ .

The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint $(x_{m}, y_{m})$ of the segment with endpoints and can be found by using the midpoint equations, setting $x_{1} = y_{2}= 7, x_{2} = y_{1}= 0$ :

$x_{m} = \frac{x_{1}+x_{2}}{2} = \frac{7+0}{2} = \frac{7}{2}$

$y_{m} = \frac{y_{1}+y_{2}}{2} = \frac{0+7 }{2} = \frac{7}{2}$

The circle passed though this midpoint $\left ( \frac{7}{2}, \frac{7}{2} \right )$ . The segment from this point to the origin is a radius, and its length is equal to . Using the following form of the distance formula, since we only need the square of the radius:

$r ^{2} = (x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}$ ,

set $x_{1} = y_{1} = 0, x_{2}= y_{2}= \frac{7}{2}$ :

$r ^{2} =\left (\frac{7}{2}-0\right ) ^{2}+\left (\frac{7}{2}-0\right ) ^{2}$

$=\left (\frac{7}{2} \right ) ^{2}+\left (\frac{7}{2}\right ) ^{2}$

$= \frac{49}{4} + \frac{49}{4}$

$= \frac{49}{2}$

Substituting in the circle equation for $r ^{2}$ , we get the correct response,

$x^{2}+ y^{2}= \frac{49}{2}$

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Question

Find the diameter of the circle with the equation .

Answer

Start by putting the equation in the standard form of the equation of a circle by completing the square. Recall the standard form of the equation of a circle:

, where the center of the circle is at and the radius is .

From the equation, we know that .

$r=\sqrt{16}=4$

Since the radius is , double its length to find the length of the diameter. The length of the diameter is .

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