How to find the equation of a circle

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SSAT Upper Level Quantitative › How to find the equation of a circle

Questions 1 - 10

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 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:

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

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

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 circumscribed about the square.

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

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

$(x-3)^{2}+(y-3)^{2} = 36$

$(x-6)^{2}+(y-6)^{2} = 18$

$(x-6)^{2}+(y-6)^{2} = 36$

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 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 $r = 3 \sqrt{2}$ .

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

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

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

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

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

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

$2 \pi \sqrt{11}$

$22 \pi$

$2 \pi$

$2 \pi \sqrt{2}$

$\pi \sqrt{13}$

Explanation

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

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

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

None of the other responses gives the correct answer.

$x^{2}+ y^{2} = 64$

$x^{2}+ y^{2} = 32$

$x^{2}+ y^{2} = 128$

$x^{2}+ y^{2} = 16$

Explanation

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.

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,