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# Digital SAT Math Practice Questions -Advanced : Circles

## SAT MAth Practice questions – all topics

• Geometry and Trigonometry Weightage: 15% Questions: 5-7
• Area and volume
• Lines, angles, and triangles
• Right triangles and trigonometry
• Circles

## SAT MAth and English  – full syllabus practice tests

[Calc]  Question  Hard

Line k is tangent to the circle with center O at point P as shown. What is the slope of line k ?

Ans:3/4

Lets denotes slope of line $\rm{OP}$ is $m$

$m = \frac{{y_2 – y_1}}{{x_2 – x_1}}$
$m = \frac{{0 – (-4)}}{{0- 3}}$
$m = \frac{4}{{-3}}$
$m = -\frac{4}{3}$

If the slope of line $$OP$$ is $$-\frac{4}{3}$$, then the slope of line $$k$$, denoted as $$m_k$$, will be the negative reciprocal of $$-\frac{4}{3}$$.

The negative reciprocal of $$-\frac{4}{3}$$ is obtained by flipping the fraction and changing the sign:
$m_k = \frac{3}{4}$

So, the slope of line $$k$$ is $$\frac{3}{4}$$.

[Calc]  Question   Hard

A circle in the $$x y$$-plane has its center at $$(-3,4)$$ and the point $$(-2,1)$$ lies on the circle. Which equation represents this circle?

A. $$(x-3)^2+(y+4)^2=\sqrt{10}$$
B. $$(x+3)^2+(y-4)^2=\sqrt{10}$$
C. $$(x-3)^2+(y+4)^2=10$$
D. $$(x+3)^2+(y-4)^2=10$$

Ans:D

To find the equation representing the circle, we’ll use the standard form of the equation for a circle:

$(x – h)^2 + (y – k)^2 = r^2$

where $$(h, k)$$ are the coordinates of the center of the circle, and $$r$$ is the radius of the circle.

We can use the distance formula to find the radius of the circle, which is the distance from the center $$(-3, 4)$$ to the point $$(-2, 1)$$:

$r = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

$r = \sqrt{(-2 – (-3))^2 + (1 – 4)^2}$

$r = \sqrt{(1)^2 + (-3)^2}$

$r = \sqrt{1 + 9}$

$r = \sqrt{10}$

So, the radius of the circle is $$\sqrt{10}$$.

Now, we can plug the center and the radius into the standard form equation of the circle:

$(x + 3)^2 + (y – 4)^2 = (\sqrt{10})^2$

$(x + 3)^2 + (y – 4)^2 = 10$

Therefore, the equation representing this circle is:

$(x + 3)^2 + (y – 4)^2 = 10$

[Calc]  Question   Hard

In the figure shown, point $$B$$ and the center of each circle lie on $$\overline{A C}$$. The ratio of $$A B$$ to $$B C$$ is 4 to 1 . If the area of the small circle is 72 , what is the area of the shaded region?

Ans:40.5,81 / 2

$\rm{A B: B C=4: 1}$

Let’s assume that the length of $$A B$$ is $$4 a$$ and the length of $$B C$$ is $$a$$

Area of the smaller circle
\begin{aligned} & =\pi(2 a)^2 \\ & 4 \pi a^2=72 \rightarrow \pi a^2=18 \end{aligned}

The radius of the larger circle is 5a divided by 2.

Area of the larger circle
$=\pi\left(\frac{5}{2} a\right)^2$

$\text{Area of the shaded circular region = Area of the larger circle -Area of the small circle}$

\begin{aligned} & \text{Area of the shaded circular region}\\&=\pi \times \frac{25}{4} a^2-\pi \times 4 a^2 \\ & =\pi \frac{9}{4} a^2\end{aligned}

substituting $$\pi a^2=18$$

$$\text{Area of the shaded circular region}=\frac{81}{2}$$

[No calc]  Question  Hard

$$x^{2}-10x+y^{2}-6y=30$$

In the xy-plane, the graph of the given equation is a circle. What is the area of this circle?
A. 8$$\pi$$
B. 15$$\pi$$
C. 46$$\pi$$
D. 64$$\pi$$

Ans: D

To find the area of the circle represented by the equation $$x^2-10 x+y^2-6 y=30$$, we need to rewrite the equation in standard form by completing the square.

$(x-5)^2+(y-3)^2-34=30$

$(x-5)^2+(y-3)^2=64$

Standard Form of a Circle:
The standard form of a circle’s equation is $$(x-h)^2+(y-k)^2=r^2$$, where $$(h, k)$$ is the center and $$r$$ is the radius.

From $$(x-5)^2+(y-3)^2=64$$, we identify:
Center: $$(5,3)$$
Radius: $$r=\sqrt{64}=8$$

The area $$A$$ of a circle is given by $$A=\pi r^2$$.

Substitute $$r=8$$ :
$A=\pi\left(8^2\right)=\pi \cdot 64=64 \pi$

[Calc]  Question   Hard

Point P is the center of the circle in the figure shown. What is the value of x?

100

Angle subtended by the arc at circumference$$=\frac{1}{2}$$ Angle subtended by the are at the circle

\begin{aligned} \angle BAC & =\frac{1}{2} \angle BPC \\ & =\frac{1}{2} x \\ \angle BAC & =x / 2\end{aligned}

In quadrilateral ABPC Sum of all angles is 360.,

$\angle A+ \angle B + \angle P + \angle C = 360^{\circ}$

$\angle P = 360 – x$

$\frac{1}{2} x+50+360-x=360$

$x=100$

[No calc]  Question  Hard

The perimeter of a square inscribed in a circle is 30 inches. The radius of the circle is $$x\sqrt{2}$$ inches. What is the value of x?

Ans: 15/4, 3.75

Let $$s$$ be the side length of the square, and $$r$$ be the radius of the circle. The perimeter of the square is $$4s$$, and the diameter of the circle (which is equal to twice the radius) is equal to the diagonal of the square. Since the diagonal of a square is $$s \sqrt{2}$$, and it’s also the diameter of the circle, we have:

$s \sqrt{2} = 2r$

Given that the perimeter of the square is $$30$$ inches, we know $$4s = 30$$, so $$s = \frac{30}{4} = 7.5$$ inches.

Now, we’ll substitute the given value of $$r = x \sqrt{2}$$ into the equation:

$7.5 \sqrt{2} = 2(x \sqrt{2})$
$7.5 = 2x$
$x = \frac{7.5}{2}$
$x = 3.75$

So, the value of $$x$$ is $$3.75$$.

[Calc]  Question  Hard

The circle shown has center $$(-1,1)$$. Line $$t$$ is tangent to this circle at point $$(4,-3)$$. Which of the following points also lies on line $$t$$ ?
A) $$\left(0, \frac{5}{4}\right)$$
B) $$(3,6)$$
C) $$(8,2)$$
D) $$(9,1)$$

C

\begin{aligned} m & =\frac{y_2-y_1}{x_2-x_1} \\ & =\frac{-3-1}{4-(-1)} \\ m & =-\frac{4}{5}\end{aligned}

Now $\frac{-1}{m}$ will slope of t.

Slope of t $= \frac{5}{4}$ and line T is passing from $$(4,-3)$$.

To find the equation of the line $$t$$ that passes through the point $$(4, -3)$$ with a slope of $$\frac{5}{4}$$, we can use the point-slope form of the equation of a line:

$y – y_1 = m(x – x_1)$

Here, $$(x_1, y_1)$$ is the point $$(4, -3)$$ and $$m$$ is the slope $$-\frac{4}{5}$$. Substituting these values in:

$y – (-3) = \frac{5}{4}(x – 4)$

$y + 3 = \frac{5}{4}(x – 4)$

\begin{aligned} & y+3=\frac{5}{4} x-\frac{5 \cdot 4}{4} \\ & y+3=\frac{5}{4} x-5 \end{aligned}

Next, subtract 3 from both sides to isolate $$y$$ :
$y=\frac{5}{4} x-5-3$

Combine the constants on the right-hand side:
$y=\frac{5}{4} x-8$

To determine which of the following points lies on the line $$t$$ with the equation $$y = \frac{5}{4}x – 8$$, we need to substitute each point into the equation and check if the equation holds true.

A) $$\left(0, \frac{5}{4}\right)$$:

Substitute $$x = 0$$ and $$y = \frac{5}{4}$$:

$y = \frac{5}{4}(0) – 8 = -8$

Since $$\frac{5}{4} \neq -8$$, this point does not lie on the line.

B) $$(3, 6)$$:

Substitute $$x = 3$$ and $$y = 6$$:

$y = \frac{5}{4}(3) – 8$
$y = \frac{15}{4} – 8$
$y = \frac{15}{4} – \frac{32}{4}$
$y = \frac{-17}{4}$

Since $$6 \neq \frac{-17}{4}$$, this point does not lie on the line.

C) $$(8, 2)$$:

Substitute $$x = 8$$ and $$y = 2$$:

$y = \frac{5}{4}(8) – 8$
$y = 10 – 8$
$y = 2$

Since $$2 = 2$$, this point lies on the line.

D) $$(9, 1)$$:

Substitute $$x = 9$$ and $$y = 1$$:

$y = \frac{5}{4}(9) – 8$
$y = \frac{45}{4} – 8$
$y = \frac{45}{4} – \frac{32}{4}$
$y = \frac{13}{4}$

Since $$1 \neq \frac{13}{4}$$, this point does not lie on the line.

Therefore, the point $$(8, 2)$$ lies on the line $$t$$. The correct answer is:

C) $$(8, 2)$$

[No calc]  Question   Hard

$$x^{2}-10x+y^{2}-6y-47=0$$

In the xy-plane, the graph of the given equation is a circle. If this circle is inscribed in a square, what is the perimeter of the square?
A. 18
B. 36
C. 72
D. 188

Ans: C

To find the perimeter of the square inscribed in the circle represented by the given equation $$x^2 – 10x + y^2 – 6y – 47 = 0$$, we first need to determine the radius of the circle.

The given equation represents a circle because it follows the general form of the equation for a circle:
$(x – h)^2 + (y – k)^2 = r^2$

Where $$(h, k)$$ is the center of the circle and $$r$$ is the radius.

To find the center of the circle, we complete the square for both the $$x$$-terms and $$y$$-terms:
$x^2 – 10x + y^2 – 6y – 47 = 0$
$(x^2 – 10x + 25) + (y^2 – 6y + 9) – 47 – 25 – 9 = 0$
$(x – 5)^2 + (y – 3)^2 = 81$

Comparing this to the standard form $$(x – h)^2 + (y – k)^2 = r^2$$, we see that the center of the circle is $$(5, 3)$$ and the radius is $$\sqrt{81} = 9$$.

Now, if a circle is inscribed in a square, each side of the square is equal to twice the radius of the circle.

$a= 2r$

So, the perimeter of the square is $$4 \times \text{radius} \times 2$$, which is $$4 \times 9 \times 2 = 72$$.

Therefore, the correct answer is option C) 72.

[Calc]  Question Hard

A circle in the xy-plane has its center at (4, −5) and has a radius of 2. An equation of the circle is  $$x^{2}-8x+y^{2}+10y+C=0$$, where c is a constant. What is the value of c?

Ans: 37

$(x-h)^2+(y-k)^2=r^2$
where $$(h, k)$$ is the center of the circle and $$r$$ is the radius.

Given that the center of the circle is $$(4,-5)$$ and the radius is 2 , we substitute these values into the standard form:
\begin{aligned} & (x-4)^2+(y+5)^2=2^2 \\ & (x-4)^2+(y+5)^2=4 \end{aligned}

Now, we expand the left side of the equation:
\begin{aligned} & x^2-8 x+16+y^2+10 y+25=4 \\ & x^2-8 x+y^2+10 y+41=4 \\ & x^2-8 x+y^2+10 y+41-4=0 \\ & x^2-8 x+y^2+10 y+37=0 \end{aligned}

Comparing this with the given equation $$x^2-8 x+y^2+10 y+C=0$$, we see that $$C=37$$.

[No calc]  Question  Hard

$$x^2+y^2+6x+5y=\frac{-45}{4}$$

The equation of a circle in the xy-plane is shown. What is the radius of the circle?

2

Given the equation of a circle in the $$xy$$-plane:
$x^2 + y^2 + 6x + 5y = -\frac{45}{4}$

To find the radius, complete the square for both $$x$$ and $$y$$.

Rewrite the equation grouping $$x$$ and $$y$$ terms:
$x^2 + 6x + y^2 + 5y = -\frac{45}{4}$

Complete the square for $$x$$:
$x^2 + 6x \rightarrow (x + 3)^2 – 9$

Complete the square for $$y$$:
$y^2 + 5y \rightarrow (y + \frac{5}{2})^2 – \left(\frac{5}{2}\right)^2 = (y + \frac{5}{2})^2 – \frac{25}{4}$

Substitute these into the equation:
$(x + 3)^2 – 9 + (y + \frac{5}{2})^2 – \frac{25}{4} = -\frac{45}{4}$

Combine constants on the right side:
$(x + 3)^2 + (y + \frac{5}{2})^2 = -\frac{45}{4} + 9 + \frac{25}{4}$
$(x + 3)^2 + (y + \frac{5}{2})^2 = -\frac{45}{4} + \frac{36}{4} + \frac{25}{4}$
$(x + 3)^2 + (y + \frac{5}{2})^2 = \frac{16}{4} = 4$

Thus, the equation of the circle is:
$(x + 3)^2 + (y + \frac{5}{2})^2 = 4$

The radius $$r$$ of the circle is the square root of 4:
$r = \sqrt{4} = 2$

[Calc]  Question  Hard

What is the diameter of the circle in the xy-plane with equation $$(x-5)^{2}+(y-4)^{2}=64$$?

Ans: 16

The equation of a circle in the form $$(x – h)^2 + (y – k)^2 = r^2$$ reveals the center $$(h, k)$$ and the radius $$r$$.

Given:
$(x – 5)^2 + (y – 4)^2 = 64,$

$(h, k) = (5, 4) \quad \text{and} \quad r^2 = 64 \Rightarrow r = \sqrt{64} = 8.$

The diameter $$D$$ of the circle is:
$D = 2r = 2 \times 8 = 16.$

[Calc]  Question  Hard

In the figure shown, point O is the center of the circle. One vertex of the square lies on the circle, and the opposite vertex is point O. If the area of the shaded region is 36$$\pi$$ – 18, what is the perimeter of the square?
A) 24
B) 72
C) $$12\sqrt{2}$$
D) $$36\sqrt{2}$$

Ans: C

$\sqrt{2}S=R$

Area of Square $= \text{side}^2$

Area of Square $=(\frac{R}{\sqrt{2}})^2\Rightarrow \frac{R^2}{2}$

Area of Circle $= \pi \text{R}^2$

$\text{area of the shaded region= Area of Circle – Area of Square}$

$\text{area of the shaded region}\Rightarrow R^2(\pi -\frac{1}{2})=36 (\pi -\frac{1}{2})$

From this $R= 6$

perimeter of the square = 4S

perimeter of the square $= 4\times \frac{R}{\sqrt{2}}$

perimeter of the square $= 4\times \frac{6}{\sqrt{2}}\Rightarrow 12\sqrt{2}$

[No- Calc]  Question   Hard

$x^2-6 x+y^2-8 y=0$

The graph of the given equation in the $$x y$$-plane is a circle. What is the radius of the circle?

A) 2
B) 3
C) 4
D) 5

Ans:D

To find the radius of the circle from the equation $$x^2 – 6x + y^2 – 8y = 0$$, we need to rewrite it in the standard form of a circle equation $$(x-h)^2 + (y-k)^2 = r^2$$ by completing the square.

Starting with the given equation:
$x^2 – 6x + y^2 – 8y = 0$

Complete the square for $$x$$:
$x^2 – 6x$
$= (x – 3)^2 – 9$

Complete the square for $$y$$:
$y^2 – 8y$
$= (y – 4)^2 – 16$

Now substitute these completed squares back into the equation:
$(x – 3)^2 – 9 + (y – 4)^2 – 16 = 0$

Combine and simplify:
$(x – 3)^2 + (y – 4)^2 – 25 = 0$
$(x – 3)^2 + (y – 4)^2 = 25$

This is the standard form of a circle equation where $$(h, k) = (3, 4)$$ and $$r^2 = 25$$. Therefore, the radius $$r = \sqrt{25} = 5$$.

D) 5

[Calc]  Question  Hard

Points A and B lie on a circle with radius 4 meters, and $\operatorname{arc} \overline{A B}$ has length meters. The length $\frac{4 \pi}{5}$ of $\operatorname{\operatorname {arc}} \overline{A B}$ is what fraction of the circumference of the circle?

$1 / 10, .1$

[Calc]  Question  Hard

In the circle shown, point $O$ is the center, and diameter $\overline{U X}$ bisects $\overline{W Y}$ at point $\mathrm{Z}$. The radius of the circle is 6 , and $X Z=2$. If $W Z=\sqrt{x}$, what is the value of $x$ ?

20

[Calc]  Question  Hard

A square is inscribed in a circle with radius $6 \sqrt{2}$ inches. What is the perimeter of the square in inches?

48

Question

$$(x + 3)^{2} + (y – 7)^{2} = 100$$

In the $$xy$$-plane, the graph of the given equation is a circle. Which point $$(x, y)$$ lies on the circle?

1. (3, -4)
2. (3, -1)
3. (3, 1)
4. (3, 4)

B

Questions

$(x-6)^2+(y-3)^2=25$

The graph in the $x y$-plane of the equation above is a circle. If the circle is translated downwards $a$ units such that the circle is tangent to the $x$ axis, the equation becomes $(x-6)^2+(y-3+a)^2=25$. What is the value of $a$ ?

Ans: 8

Questions

The equation $(x+6)^2+(y+3)^2=121$ defines a circle in the $x y$-plane. What is the radius of the circle?

Ans: 11

Questions

A circle in the $x y$-plane has a diameter with endpoints $(-1,-3)$ and $(7,3)$. If the point $(0, b)$ lies on the circle and $b>0$, what is the value of $b$

Ans: 4

Questions

An arc of a circle measures 2.4 radians. To the nearest degree, what is the measure, in degrees, of this arc? (Disregard the degree sign when gridding your answer.)

5Ans: 138,137

Questions

Segments $$\bar{OA}$$ and $$\bar{OA}$$ are radii of the semi circle above. Arc $$\widehat{AB}$$ has length $$3\pi$$ and $$OA=5$$. What is the value of $$x$$?

Ans: 108

Question

Which of the following equations describes a circle with radius 10 that passes through the origin when graphed in the $x y$-plane?
A. $(x-5)^2+(y+5)^2=10$
B. $(x-5)^2+(y+5)^2=100$
C. $(x-10)^2+(y-10)^2=100$
D. $(x-5 \sqrt{2})^2+(y+5 \sqrt{2})^2=100$

Ans: D

Question

The graph of $x^2-4 x+y^2+6 y-24=0$ in the $x y$-plane is a circle. What is the radius of the circle?
A. $2 \sqrt{6}$
B. $\sqrt{11}$
C. $\sqrt{37}$
D. $\sqrt{76}$

Ans: C

Question

In the figure above, the circle has center $$A$$, and line segment $$CB$$ is tangent to the circle at point $$C$$. If $$AB$$ = 1.0 and $$CB$$ = 0.8, what is the length of the diameter of the circle?

Ans: 1.2, 6/5

Question

An instrument shows the number of revolutions per minute made by each tire of a car. In each revolution, the car travels a distance equal to the circumference of one of its tires. The circumference of each tire is equal to $2 \pi r$, where $r$ is the radius of the tire.
Maria gets new tires for her car. The radius of each of her old tires is 0.30 meter, and the radius of each of her new tires is $11 \%$ larger than the radius of one of her old tires. What is the circumference of each new tire, to the nearest tenth of a meter?

Ans: $2.1,21 / 10$

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