Circles - SSAT Upper Level Quantitative

Card 0 of 56

Question

Give the equation of the above circle.

Answer

A circle with center and radius has equation

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

The circle has center and radius 4, so substitute:

$(x-(-3))^{2}+ (y-2)^{2}= 4^{2}$

$(x+3)^{2}+ (y-2)^{2}= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4- 16= 16-16$

$x^{2}+6x + y ^{2}-4y-3=0$

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Question

Give the equation of the above circle.

Answer

A circle with center and radius has equation

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

The circle has center and radius 5, so substitute:

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

$(x-2)^{2}+ (y-4)^{2}= 25$

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Question

A circle on the coordinate plane has a diameter whose endpoints are and . Give its equation.

Answer

A circle with center and radius has equation

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

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{-3+9}{2}, \frac{7+5}{2} \right )$

Therefore, and

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using and . We will concern ourcelves with finding the square of the radius $r^{2}$ :

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

$r^{2}= \left (9-3\right )^{2} + \left ( 5-6\right )^{2}$

$r^{2}=6^{2} + \left (-1\right )^{2}$

$r^{2}=36+ 1= 37$

Substitute:

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

$(x-3)^{2}+ (y-6)^{2}=37$

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Question

A circle on the coordinate plane has a diameter whose endpoints are and . Give its equation.

Answer

A circle with center and radius has equation

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

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{2+(-8)}{2}, \frac{7+3}{2} \right )$

Therefore, and .

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using and . We will concern ourcelves with finding the square of the radius $r^{2}$ :

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

$r^{2}= \left ( 2- (-3)\right )^{2} + \left ( 7-5\right )^{2}$

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

$r^{2}=25+ 4 = 29$

Substitute:

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

$(x-(-3))^{2}+ (y-5)^{2}= 29$

$(x+3)^{2}+ (y-5)^{2}= 29$

Expand:

$x^{2}+6x+9+ y^{2}-10y+25= 29$

$x^{2}+6x + y^{2}-10y+34= 29$

$x^{2}+6x + y^{2}-10y+34- 29 = 0$

$x^{2}+6x + y^{2}-10y+5 = 0$

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Question

A circle on the coordinate plane has center and circumference $20 \pi$ . Give its equation.

Answer

A circle with center and radius has equation

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

The center is , so .

To find , use the circumference formula:

$2 \pi r = C$

$2 \pi r = 20 \pi$

$r^{2} = 100$

Substitute:

$(x-4)^{2}+ (y-(-6))^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=100$

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Question

A circle on the coordinate plane has center and area $36 \pi$ . Give its equation.

Answer

A circle with center and radius has the equation

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

The center is , so .

The area is $36 \pi$ , so to find $r^{2}$ , use the area formula:

$\pi r^{2}= A$

$\pi r^{2}= 36 \pi$

$r^{2}= 36$

The equation of the line is therefore:

$(x-2)^{2}+ (y-5)^{2}= 36$

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Question

What is the equation of a circle that has its center at and has a radius of ?

Answer

The general equation of a circle with center and radius is:

Now, plug in the values given by the question:

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Question

If the center of a circle with a diameter of 5 is located at , what is the equation of the circle?

Answer

Write the formula for the equation of a circle with a given point, .

The radius of the circle is half the diameter, or .

Substitute all the values into the formula and simplify.

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Question

Give the circumference of the circle on the coordinate plane whose equation is

$x^{2}+ y^{2}+ 4x- 6y + 2 = 0$

Answer

The standard form of the equation of a circle is

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

where is the radius of the circle.

We can rewrite the equation we are given, which is in general form, in this standard form as follows:

$x^{2}+ y^{2}+ 4x- 6y + 2 = 0$

$x^{2}+ y^{2}+ 4x- 6y = -2$

$\left (x^{2}+ 4x \right )+\left ( y^{2}- 6y \right )= -2$

Complete the squares. Since $\left ( \frac{4}{2} \right )^{2} = 4$ and $\left ( \frac{-6}{2} \right )^{2} = 9$ , we do this as follows:

$\left (x^{2}+ 4x +4 \right )+\left ( y^{2}- 6y +9\right )= -2+4+9$

$(x+2)^{2}+(y-3)^{2} = 11$

$r^{2} = 11$ , so $r = \sqrt{11}$ , and the circumference of the circle is

$C = 2 \pi r = 2 \pi \sqrt{11}$

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Question

Which of the following is the equation of a circle with center at the origin and circumference $64 \pi$ ?

Answer

The standard form of the equation of a circle is

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

where the center is and the radius is .

The center of the circle is the origin, so .

The equation will be

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

for some .

The circumference of the circle is $64 \pi$ , so

$C = 2 \pi r = 64 \pi$

$\frac{2 \pi r }{2 \pi }= \frac{64 \pi}{2 \pi }$

$r^{2}= 32^{2} = 1,024$

The equation is $x^{2}+ y^{2} = 1,024$ , which is not among the responses.

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Question

Which of the following is the equation of a circle with center at the origin and area $64 \pi$ ?

Answer

The standard form of the equation of a circle is

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

where the center is and the radius is .

The center of the circle is the origin, so , and the equation is

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

for some .

The area of the circle is $64 \pi$ , so

$A = \pi r^{2} = 50 \pi$

$\frac{ \pi r^{2}}{\pi} = \frac{ 64 \pi}{\pi}$

$r^{2} = 64$

We need go no further; we can substitute to get the equation $x^{2}+ y^{2} = 64$ .

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Question

A square on the coordinate plane has as its vertices the points . Give the equation of a circle circumscribed about the square.

Answer

Below is the figure with the circle and square in question:

The center of the inscribed circle coincides with that of the square, which is the point . Its diameter is the length of a diagonal of the square, which is $\sqrt{2}$ times the sidelength 6 of the square - this is $6 \sqrt{2}$ . Its radius is, consequently, half this, or $3 \sqrt{2}$ . Therefore, in the standard form of the equation,

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

substitute and $r = 3 \sqrt{2}$ .

$(x-3)^{2}+(y-3)^{2} = \left (3 \sqrt{2} \right ) ^{2}$

$(x-3)^{2}+(y-3)^{2} = 3^{2} \left ( \sqrt{2} \right ) ^{2}$

$(x-3)^{2}+(y-3)^{2} = 9 \cdot 2$

$(x-3)^{2}+(y-3)^{2} = 18$

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Question

A square on the coordinate plane has as its vertices the points . Give the equation of a circle inscribed in the square.

Answer

Below is the figure with the circle and square in question:

The center of the inscribed circle coincides with that of the square, which is the point . Its diameter is equal to the sidelength of the square, which is 8, so, consequently, its radius is half this, or 4. Therefore, in the standard form of the equation,

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

substitute and .

$(x-4)^{2}+(y-4)^{2} = 4^{2}$

$(x-4)^{2}+(y-4)^{2} = 16$

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Question

Give the area of the circle on the coordinate plane whose equation is

$x^{2}+ y^{2}+ 4x- 8y -11 = 0$ .

Answer

The standard form of the equation of a circle is

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

where is the radius of the circle.

We can rewrite the equation we are given, which is in general form, in this standard form as follows:

$x^{2}+ y^{2}+ 4x- 8y -11 = 0$

$x^{2}+ y^{2}+ 4x- 8y =11$

$\left (x^{2}+ 4x \right ) +\left ( y^{2}- 8y \right ) =11$

Complete the squares. Since $\left ( \frac{4}{2} \right )^{2} = 4$ and $\left ( \frac{-8}{2} \right )^{2} = 16$ , we do this as follows:

$\left (x^{2}+ 4x + 4 \right ) +\left ( y^{2}- 8y + 16\right ) =11 + 4 + 16$

$(x-2)^{2}+(y-4)^{2} = 31$

$r^{2}= 31$ , and the area of the circle is

$A = \pi r^{2} = 31 \pi$

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Question

Give the equation of the above circle.

Answer

A circle with center and radius has equation

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

The circle has center and radius 4, so substitute:

$(x-(-3))^{2}+ (y-2)^{2}= 4^{2}$

$(x+3)^{2}+ (y-2)^{2}= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4- 16= 16-16$

$x^{2}+6x + y ^{2}-4y-3=0$

Compare your answer with the correct one above

Question

Give the equation of the above circle.

Answer

A circle with center and radius has equation

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

The circle has center and radius 5, so substitute:

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

$(x-2)^{2}+ (y-4)^{2}= 25$

Compare your answer with the correct one above

Question

A circle on the coordinate plane has a diameter whose endpoints are and . Give its equation.

Answer

A circle with center and radius has equation

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

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{-3+9}{2}, \frac{7+5}{2} \right )$

Therefore, and

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using and . We will concern ourcelves with finding the square of the radius $r^{2}$ :

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

$r^{2}= \left (9-3\right )^{2} + \left ( 5-6\right )^{2}$

$r^{2}=6^{2} + \left (-1\right )^{2}$

$r^{2}=36+ 1= 37$

Substitute:

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

$(x-3)^{2}+ (y-6)^{2}=37$

Compare your answer with the correct one above

Question

A circle on the coordinate plane has a diameter whose endpoints are and . Give its equation.

Answer

A circle with center and radius has equation

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

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{2+(-8)}{2}, \frac{7+3}{2} \right )$

Therefore, and .

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using and . We will concern ourcelves with finding the square of the radius $r^{2}$ :

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

$r^{2}= \left ( 2- (-3)\right )^{2} + \left ( 7-5\right )^{2}$

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

$r^{2}=25+ 4 = 29$

Substitute:

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

$(x-(-3))^{2}+ (y-5)^{2}= 29$

$(x+3)^{2}+ (y-5)^{2}= 29$

Expand:

$x^{2}+6x+9+ y^{2}-10y+25= 29$

$x^{2}+6x + y^{2}-10y+34= 29$

$x^{2}+6x + y^{2}-10y+34- 29 = 0$

$x^{2}+6x + y^{2}-10y+5 = 0$

Compare your answer with the correct one above

Question

A circle on the coordinate plane has center and circumference $20 \pi$ . Give its equation.

Answer

A circle with center and radius has equation

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

The center is , so .

To find , use the circumference formula:

$2 \pi r = C$

$2 \pi r = 20 \pi$

$r^{2} = 100$

Substitute:

$(x-4)^{2}+ (y-(-6))^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=100$

Compare your answer with the correct one above

Question

A circle on the coordinate plane has center and area $36 \pi$ . Give its equation.

Answer

A circle with center and radius has the equation

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

The center is , so .

The area is $36 \pi$ , so to find $r^{2}$ , use the area formula:

$\pi r^{2}= A$

$\pi r^{2}= 36 \pi$

$r^{2}= 36$

The equation of the line is therefore:

$(x-2)^{2}+ (y-5)^{2}= 36$

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