Circles

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SSAT Upper Level Quantitative › Circles

Questions 1 - 10

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

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

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

$(x+4)^{2}+ (y-6)^{2}=10$

$(x-4)^{2}+ (y+6)^{2}=10$

$(x+4)^{2}+ (y-6)^{2}=400$

Explanation

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:

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

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

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

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

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

$(x+4)^{2}+ (y-6)^{2}=10$

$(x-4)^{2}+ (y+6)^{2}=10$

$(x+4)^{2}+ (y-6)^{2}=400$

Explanation

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:

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

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

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

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

$(x+2)^{2}+ (y+5)^{2}= 6$

$(x+2)^{2}+ (y+5)^{2}= 18$

$(x-2)^{2}+ (y-5)^{2}= 6$

$(x-2)^{2}+ (y-5)^{2}=18$

Explanation

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:

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

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

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

$(x+2)^{2}+ (y+5)^{2}= 6$

$(x+2)^{2}+ (y+5)^{2}= 18$

$(x-2)^{2}+ (y-5)^{2}= 6$

$(x-2)^{2}+ (y-5)^{2}=18$

Explanation

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:

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

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

Explanation

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

Now, plug in the values given by the question:

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

Explanation

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

Now, plug in the values given by the question:

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

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

$(x-4)^{2}+(y-4)^{2} = 32$

$(x-8)^{2}+(y-8)^{2} = 16$

$(x-8)^{2}+(y-8)^{2} = 32$

$(x-8)^{2}+(y-8)^{2} = 64$

Explanation

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

Circle on axes

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,