Statement 1 : The only circle having radius $\sqrt{10}$ and a diameter along line $2 x+y=5 \, is\, x^{2}+y^{2}-6x+2y=0.$ Statement 2 : $2x + y = 5$ is a normal to the circle $x^{2}+y^{2}-6x+2y=0.$

Question

The area bounded by $ y - 1 = |x| $ and $ y + 1 = |x| $ is: (a) $ \frac{1}{2} $

Answer 1

To determine the truthfulness of the given statements, let's analyze each step-by-step.

Consider the equation of the circle $ x^{2} + y^{2} - 6x + 2y = 0 $.
First, we write the given circle equation in standard form by completing the square:
- Rearrange terms: $ (x^2 - 6x) + (y^2 + 2y) = 0 $.
- Complete the square for $x$:
  - $ x^2 - 6x = (x - 3)^2 - 9 $.
- Complete the square for $y$:
  - $ y^2 + 2y = (y + 1)^2 - 1 $.
- Substitute these back into the equation:
  - $(x - 3)^2 - 9 + (y + 1)^2 - 1 = 0$.
  - Simplify: $(x - 3)^2 + (y + 1)^2 = 10$.
This is the equation of a circle with center $(3, -1)$ and radius $\sqrt{10}$.
Check if the diameter lies on the line $2x + y = 5$:
- The line is $(y = 5 - 2x)$. We substitute a general point $ (h, k) $ on this line.
- The center $(3, -1)$ is used to check if the line is a diameter:
- Line's normality condition with center: If the radius is perpendicular to the line, the slope product should be -1.
- The slope of $2x + y = 5$ is $ -2 $, while the radius direction $(x - 3, y + 1)$ becomes zero parallel to the diameter constraint, invalidating the statement.
Check if the line is a normal to the circle:
- Since the line $2x + y = 5$ is perpendicular to any line passing through the center and satisfying the normal condition, Statement 2 holds true.
- Thus, $2x + y = 5$ is indeed a normal to the circle.

Hence, the conclusion is:

Statement 1 is false because the line $2x + y = 5$ does not act as a diameter constraint on the circle's alignment with $ (3, -1) $.
Statement 2 is true because the line meets the criteria of being normal to the circle.

Therefore, the correct answer is: Statement 1 is false; Statement 2 is true.

The Correct Option is A