Q. A circle with the equation \( (x – h)^2 + (y – k)^2 = r^2 \) is shifted 3 units to the right and 2 units down. What is the new equation of the circle?
Solution:
The original center of the circle is
\( (h, k) \).
After shifting 3 units to the right and 2 units down, the new center becomes:
\( (h + 3, \, k – 2) \).
The radius remains unchanged, so it is still \( r \).
New equation of the circle:
\( (x – (h + 3))^2 + (y – (k – 2))^2 = r^2 \)
Question: If the line \( y = mx + c \) is reflected in the x-axis, find the equation of the reflected line.
Solution:
To reflect a line in the x-axis, we replace \( y \) with \( -y \).
So, starting with:
\( y = mx + c \)
Replace \( y \) by \( -y \):
\( -y = mx + c \)
Multiplying both sides by \( -1 \):
\( y = -mx – c \)
Thus, the equation of the reflected line is \( y = -mx – c \).
Question: If the line \( y = mx + c \) is reflected in the x-axis, find the equation of the reflected line.
Solution:
To reflect a line in the x-axis, we replace \( y \) with \( -y \).
So, starting with:
\( y = mx + c \)
Replace \( y \) by \( -y \):
\( -y = mx + c \)
Multiplying both sides by \( -1 \):
\( y = -mx – c \)
Thus, the equation of the reflected line is \( y = -mx – c \).
Class 11: Graphs in 2D Plane (Q&A)
Q1. A circle \(x^2 + y^2 = 25\) is translated 3 units right and 2 units down. What is the resulting equation?
Answer: The original center is \( (0,0) \). After shifting \(+3\) in x-direction and \(-2\) in y-direction, new center is \( (3, -2) \).
Equation: \( (x – 3)^2 + (y + 2)^2 = 25 \).
Q2. Find the equation of the line \(2x – 3y + 6 = 0\) after it is reflected in the x-axis.
Answer: Replace \( y \) with \( -y \):
\( 2x – 3(-y) + 6 = 0 \Rightarrow 2x + 3y + 6 = 0 \).
Q3. How many points of intersection exist between the graphs of \(y = |x|\) and \(y = 2 – x^2\)?
Answer: Solve \( |x| = 2 – x^2 \).
For \( x \ge 0 \): \( x = 2 – x^2 \Rightarrow x^2 + x – 2 = 0 \Rightarrow x = 1 \).
By symmetry, \( x = -1 \) also works.
Points: \( (1,1), (-1,1) \).
Total = 2 points.
Q4. Determine the equation of the perpendicular bisector of the segment joining \(A(1,2)\) and \(B(3,8)\).
Answer:
Midpoint \( M = (2,5) \).
Slope of \( AB = 3 \).
Perpendicular slope \( = -\frac{1}{3} \).
Equation: \( y – 5 = -\frac{1}{3}(x – 2) \Rightarrow x + 3y – 17 = 0 \).
Q5. Calculate the area of the triangle bounded by \(y = x\), \(y = -x\), and \(y = 4\).
Answer:
Vertices: \( (0,0), (4,4), (-4,4) \).
Base = 8, Height = 4.
Area \( = \frac{1}{2} \times 8 \times 4 = 16 \).
Q6. For the parabola \(y^2 = 12x\), find focus and length of latus rectum.
Answer:
Compare with \( y^2 = 4ax \Rightarrow a = 3 \).
Focus = \( (3,0) \).
Latus rectum length = \( 4a = 12 \).
Q7. What is the slope of the tangent to the circle \(x^2 + y^2 = 25\) at the point \( (3,4) \)?
Answer:
Slope of radius \( = \frac{4}{3} \).
Tangent slope \( = -\frac{3}{4} \).
Q8. Find the x-intercepts of the graph of the signum function \( f(x) = \text{sgn}(x^2 – 1) \).
Answer:
\( x^2 – 1 = 0 \Rightarrow x = \pm 1 \).
Intercepts: \( (1,0), (-1,0) \).
Q9. A point moves so that its distance from the origin is twice its distance from the y-axis. Find the equation of the path.
Answer:
\( \sqrt{x^2 + y^2} = 2|x| \Rightarrow x^2 + y^2 = 4x^2 \Rightarrow y^2 = 3x^2 \).
Q10. Sketching the greatest integer function \( y = [x] \) for \( 0 \le x < 3 \), how many horizontal segments are drawn?
Answer:
\( 0 \le x < 1 \Rightarrow y = 0 \)
\( 1 \le x < 2 \Rightarrow y = 1 \)
\( 2 \le x < 3 \Rightarrow y = 2 \)
Total segments = 3.