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
Question Medium
A circle in the \(x y\)-plane has its center at \((3,5)\) and has a radius of 6 . What is an equation of the circle?
A) \((x-3)^2+(y-5)^2=6\)
B) \((x+3)^2+(y+5)^2=6\)
C) \((x-3)^2+(y-5)^2=36\)
D) \((x+3)^2+(y+5)^2=36\)
▶️Answer/Explanation
Ans: C
The standard form for the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[
(x – h)^2 + (y – k)^2 = r^2
\]
Given the center \((3, 5)\) and radius \(6\), we substitute \(h = 3\), \(k = 5\), and \(r = 6\):
\[
(x – 3)^2 + (y – 5)^2 = 6^2
\]
Simplify \(6^2\):
\[
(x – 3)^2 + (y – 5)^2 = 36
\]
So, the correct answer is:
\[
\boxed{C}
\]
Question medium
Line \(k\) is tangent to the circle with center \(C\) at point \(A\), as shown. What is the slope of line \(k\) ?
A. -2
B. \(-\frac{1}{2}\)
C. \(\frac{1}{2}\)
D. 2
▶️Answer/Explanation
Ans:A
To find the slope of line \(k\), which is tangent to the circle at point \(A\), we need to find the slope of the line passing through points \(C\) and \(A\). The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[m = \frac{{y_2 – y_1}}{{x_2 – x_1}}\]
Given points \(C(-5, -4)\) and \(A(-7, -5)\):
\[m = \frac{{-5 – (-4)}}{{-7 – (-5)}}\]
\[m = \frac{{-1}}{{-7 + 5}}\]
\[m = \frac{{-1}}{{-2}}\]
\[m = \frac{1}{2}\]
So, the slope of line \(k\) is \(\frac{1}{2}\), which corresponds to option C.
Question medium
A circle has been divided into three nonoverlapping regions: I, II, and III. The area of region I is \(4\pi \) square centimeters (\(cm^{2}\)), the area of region II is \(12\pi \) \(cm^{2}\), and the area of region III is \(16\pi \) \(cm^{2}\). If a point in the circle is selected at random, what is the probability of selecting a point that does not lie in region II? (Express your answer as a decimal or fraction, not as a percent.)
▶️Answer/Explanation
Ans: 5/8, .625
To find the probability of selecting a point that does not lie in region II, we need to find the total area of the circle and subtract the area of region II, then divide by the total area of the circle.
Given:
Area of region I = \(4\pi \text{ cm}^2\)
Area of region II = \(12\pi \text{ cm}^2\)
Area of region III = \(16\pi \text{ cm}^2\)
The total area of the circle is the sum of the areas of all three regions:
\[4\pi + 12\pi + 16\pi = 32\pi \text{ cm}^2\]
So, the probability of selecting a point that does not lie in region II is:
\[\frac{4\pi + 16\pi}{32\pi} = \frac{20\pi}{32\pi} = \frac{5}{8}\]
Therefore, the probability is \(\frac{5}{8}\).
Question Medium
\(x^{2}-10x+y^{2}+6y=2\)
The graph in the xy-plane of the equation above is a circle. What are the coordinates of the center of the circle?
A) (-5,-3)
B) (-5,3)
C) (5,-3)
D) (5,3)
▶️Answer/Explanation
Ans: C
\(x^2-10x+y^2+6y=2\) represents a circle equation in the form \((x – h)^2 + (y – k)^2 = r^2\), where \((h, k)\) is the center of the circle.
For \(x\), we complete the square by adding \((10/2)^2 = 25\) inside the parenthesis:
\[x^2 – 10x + 25 + y^2 + 6y = 2 + 25\]
\[x^2 – 10x + 25 + y^2 + 6y + 9 = 27\]
\[(x – 5)^2 + (y + 3)^2 = 27\]
Comparing this to the standard form \((x – h)^2 + (y – k)^2 = r^2\), we see that the center of the circle is \((h, k) = (5, -3)\).
So, the coordinates of the center of the circle are C) \((5, -3)\).
Question Medium
\[
x^2+y^2-16 x-4 y+32=0
\]
In the \(x y\)-plane, the graph of the given equation is a circle. What is the length of the radius of this circle?
A) 2
B) 6
C) 8
D) 36
▶️Answer/Explanation
B
The given equation is:
\[ x^2 + y^2 – 16x – 4y + 32 = 0 \]
Rewrite the equation in standard form of a circle \((x-h)^2 + (y-k)^2 = r^2\):
First, complete the square for \(x\) and \(y\).
For \(x\):
\[ x^2 – 16x \]
Complete the square:
\[ x^2 – 16x + 64 – 64 \]
\[ (x-8)^2 – 64 \]
For \(y\):
\[ y^2 – 4y \]
Complete the square:
\[ y^2 – 4y + 4 – 4 \]
\[ (y-2)^2 – 4 \]
Rewrite the equation with the completed squares:
\[ (x-8)^2 – 64 + (y-2)^2 – 4 + 32 = 0 \]
\[ (x-8)^2 + (y-2)^2 – 36 = 0 \]
\[ (x-8)^2 + (y-2)^2 = 36 \]
The equation \((x-8)^2 + (y-2)^2 = 36\) represents a circle with radius \(\sqrt{36} = 6\).
So the answer is:
\[ \boxed{B} \]