How to find the equation of a circle - SAT Math

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Question

The endpoints of a diameter of circle A are located at points and . What is the area of the circle?

Answer

The formula for the area of a circle is given by A =πr2 . The problem gives us the endpoints of the diameter of the circle. Using the distance formula, we can find the length of the diameter. Then, because we know that the radius (r) is half the length of the diameter, we can find the length of r. Finally, we can use the formula A =πr2 to find the area.

The distance formula is

The distance between the endpoints of the diameter of the circle is:

$d=\sqrt{(6-(-2))^{2}+(8-4)^{2}}$

$d=\sqrt{8^{2}+4^{2}}$

$d=\sqrt{64+16}$

$d=\sqrt{80}=4\sqrt{5}$

To find the radius, we divide d (the length of the diameter) by two.

$r=\frac{d}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}$

Then we substitute the value of r into the formula for the area of a circle.

$A=\pi r^{2}=\pi ({2\sqrt{5}})^{2}=20\pi$

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Question

A circle is centered on point . The area of the circle is $9\pi$ . What is the equation of the circle?

Answer

The formula for a circle is

is the coordinate of the center of the circle, therefore and .

The area of a circle:

$r^2\pi = 9\pi$

Therefore:

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Question

A circle has a center at (5,5) and a radius of 2. If the format of the equation for the circle is (x-A)2+(y-B)2=C, what is C?

Answer

The circle has a center at (5,5) and a radius of 2. Therefore, the equation is (x-5)2+(y-5)2=22, or (x-5)2+(y-5)2=4.

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Question

If the center of a circle is at (0,4) and the diameter of the circle is 6, what is the equation of that circle?

Answer

The formula for the equation of a circle is:

(x-h) 2 + (y-k)2 = r2

Where (h,k) is the center of the circle.

h = 0 and k = 4

and diameter = 6 therefore radius = 3

(x-0) 2 + (y-4)2 = 32

x2 + (y-4)2 = 9

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Question

Circle A is given by the equation (x – 4)2 + (y + 3)2 = 29. Circle A is shifted up five units and left by six units. Then, its radius is doubled. What is the new equation for circle A?

Answer

The general equation of a circle is (x – h)2 + (y – k)2 = r2, where (h, k) represents the location of the circle's center, and r represents the length of its radius.

Circle A first has the equation of (x – 4)2 + (y + 3)2 = 29. This means that its center must be located at (4, –3), and its radius is √29.

We are then told that circle A is shifted up five units and then left by six units. This means that the y-coordinate of the center would increase by five, and the x-coordinate of the center would decrease by 6. Thus, the new center would be located at (4 – 6, –3 + 5), or (–2, 2).

We are then told that the radius of circle A is doubled, which means its new radius is 2√29.

Now, that we have circle A's new center and radius, we can write its general equation using (x – h)2 + (y – k)2 = r2.

(x – (–2))2 + (y – 2)2 = (2√29)2 = 22(√29)2 = 4(29) = 116.

(x + 2)2 + (y – 2)2 = 116.

The answer is (x + 2)2 + (y – 2)2 = 116.

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Question

Which of the following equations describes all the points (x, y) in a coordinate plane that are five units away from the point (–3, 6)?

Answer

We are trying to find an equation for all of the points that are the same distance (5 units) from (–3, 6). The locus of all points equidistant from a single point is a circle. In other words, we need to find an equation of a circle. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5.

The standard form of a circle is given below:

(x – h)2 + (y – k)2 = r2, where the center is located at (h, k) and r is the length of the radius.

In this case, h will be –3, k will be 6, and r will be 5.

(x – (–3))2 + (y – 6)2 = 52

(x + 3)2 + (y – 6)2 = 25

The answer is (x + 3)2 + (y – 6)2 = 25.

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Question

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an x-coordinate of two and a positive y-coordinate. What is the value of the y-coordinate?

Answer

Recall that the general form of the equation of a circle centered at the origin is:

_x_2 + _y_2 = _r_2

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

_x_2 + _y_2 = 52

_x_2 + _y_2 = 25

Now, the question asks for the positive y-coordinate when x = 2. To solve this, simply plug in for x:

22 + _y_2 = 25

4 + _y_2 = 25

_y_2 = 21

y = ±√(21)

Since our answer will be positive, it must be √(21).

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Question

What is the equation for a circle of radius 12, centered at the intersection of the two lines:

y_1 = 4_x + 3

and

y_2 = 5_x + 44?

Answer

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

4_x_ + 3 = 5_x_ + 44; 3 = x + 44; –41 = x

To find the y-coordinate, substitute into one of the equations. Let's use _y_1:

y = 4 * –41 + 3 = –164 + 3 = –161

The center of our circle is therefore: (–41, –161).

Now, recall that the general form for a circle with center at (_x_0, _y_0) is:

(x - _x_0)2 + (y - _y_0)2 = _r_2

For our data, this means that our equation is:

(x + 41)2 + (y + 161)2 = 122 or (x + 41)2 + (y + 161)2 = 144

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Question

What is the equation for a circle of radius 9, centered at the intersection of the following two lines?

Answer

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

To find the y-coordinate, substitute into one of the equations. Let's use :

$y = 2 \cdot (-6) + 14$

The center of our circle is therefore .

Now, recall that the general form for a circle with center at is

For our data, this means that our equation is:

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Question

The diameter of a circle has endpoints at points (2, 10) and (–8, –14). Which of the following points does NOT lie on the circle?

Answer

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Question

What is the radius of a circle with the equation ?

Answer

We need to expand this equation to $\dpi{100} \small x^{2}-8x+y^{2}-6y=24$ and then complete the square.

This brings us to $\dpi{100} \small x^{2}-8x+16+y^{2}-6y+9=24+16+9$ .

We simplify this to $\dpi{100} \small \left ( x-4 \right )^{2}+\left ( y-3 \right )^{2}=49$ .

Thus the radius is 7.

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Question

A circle has its origin at . The point is on the edge of the circle. What is the radius of the circle?

Answer

The radius of the circle is equal to the hypotenuse of a right triangle with sides of lengths 5 and 7.

$r=\sqrt{74}$

This radical cannot be reduced further.

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Question

A circle exists entirely in the first quadrant such that it intersects the -axis at . If the circle intersects the -axis in at least one point, what is the area of the circle?

Answer

We are given two very important pieces of information. The first is that the circle exists entirely in the first quadrant, the second is that it intersects both the - and -axis.

The fact that it is entirely in the first quadrant means that it cannot go past the two axes. For a circle to intersect the -axis in more than one point, it would necessarily move into another quadrant. Therefore, we can conclude it intersects in exactly one point.

The intersection of the circle with must also be tangential, since it can only intersect in one point. We can thus conclude that the circle must have both - and - intercepts equal to 6 and have a center of .

This leaves us with a radius of 6 and an area of:

$\pi radius^{2}=\pi6^{2} = 36\pi$

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Question

We have a square with length 2 sitting in the first quadrant with one corner touching the origin. If the square is inscribed inside a circle, find the equation of the circle.

Answer

If the square is inscribed inside the circle, in means the center of the circle is at (1,1). We need to also find the radius of the circle, which happens to be the length from the corner of the square to it's center.

Now use the equation of the circle with the center and .

We get

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Question

Find the equation of the circle centered at with a radius of 3.

Answer

Write the standard equation of a circle, where is the center of the circle, and is the radius.

Substitute the point and radius.

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Question

What is the equation of a circle with center (1,1) and a radius of 10?

Answer

The general equation for a circle with center (h, k) and radius r is given by

.

In our case, our h-value is 1 and our k-value is 1. Our r-value is 10.

Substituting each of these values into the equation for a circle gives us

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Question

The following circle is moved $\frac{\pi}{2}$ spaces to the left. Where is its new center located?

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

Answer

Remember that the general equation for a circle with center and radius is .

With that in mind, our original center is at $(\frac{\pi}{2}, \frac{3\pi}{2})$ .

If we move the center $\frac{\pi}{2}$ units to the left, that means that we are subtracting $\frac{\pi}{2}$ from our given coordinates.

$\frac{\pi}{2} - \frac{\pi}{2} = 0$

$\frac{3\pi}{2} - \frac{\pi}{2} = \frac{2\pi}{2} = \pi$

Therefore, our new center is $(0, \pi)$ .

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Question

A square on the coordinate plane has vertices at the points with coordinates , , , and . Give the equation of the circle that circumscribes the square.

Answer

The equation of the circle on the coordinate plane with radius and center is

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

The figure referenced is below:

The center of the circle is at the point of intersection of the diagonals, which, as is the case with any rectangle, bisect each other. Therefore, looking at the diagonal with endpoints and , we can set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the midpoint formula:

$h =\frac{ x_{1}+ x_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

and

$k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 9}{2} = \frac{9}{2}$

The center of the circumscribing circle is therefore $\left ( \frac{9}{2}, \frac{9}{2} \right )$ .

The radius of the circumscribing circle is the distance from this point to any point on the circle. The distance formula can be used here:

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

Since we are actually trying to find $r ^{2}$ , we will use the form

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

Choosing the radius with endpoints and $\left ( \frac{9}{2}, \frac{9}{2} \right )$ , we set $x_{1} = y_{1} = 0 , x_{2} = y_{2} = \frac{9}{2}$ and substitute:

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

$= \left ( \frac{9}{2} \right )^{2} + \left ( \frac{9}{2} \right )^{2}$

$= \frac{81}{4 } + \frac{81}{4 }$

$= \frac{81}{2}$

Setting $h =k= \frac{9}{2}$ and $r^{2}= \frac{81}{2}$ and substituting in the circle equation:

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

$\left ( x - \frac{9}{2}\right )^{2} + \left ( y- \frac{9}{2}\right )^{2} = \frac{81}{2}$ , the correct response.

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Question

A square on the coordinate plane has vertices at the points with coordinates , , , and . Give the equation of the circle that circumscribes the square.

Answer

The equation of the circle on the coordinate plane with radius and center is

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

The figure referenced is below:

The center of the circle is the origin ; the radius is 7. Therefore, setting and in the circle equation:

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

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

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

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Question

A square on the coordinate plane has vertices at the points with coordinates , , , and . Give the equation of the circle that circumscribes the square.

Answer

The equation of the circle on the coordinate plane with radius and center is

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

The figure referenced is below:

The center of the circle is at the point of intersection of the diagonals, which, as is the case with any rectangle, bisect each other. Therefore, looking at the diagonal with endpoints and , we can set $x_{1} =2, y_{1} = 0 , x_{2} =8, y_{2} = 10$ in the midpoint formula:

$h =\frac{ x_{1}+ x_{2} }{2} = \frac{2+8}{2} = \frac{10}{2} = 5$

and

$k=\frac{ y_{1}+ y_{2} }{2} = \frac{0+ 10}{2} = \frac{10}{2} = 5$

The center of the circumscribed circle is therefore .

The radius of the circle is the distance from this point to any of the vertices - we will use . The distance formula can be used here:

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

Since we are actually trying to find $r ^{2}$ , we will use the form

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

Setting $x_{1} =2, y_{1} = 0, x_{2} =5, y_{2} = 5$ :

$r^{2}= (5-2)^{2} +(5-0)^{2}$

$=3^{2} +5^{2}$

Setting and $r^{2} = 34$ in the circle equation:

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

$(x-5)^{2} + (y-5)^{2} = 34$

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