Question

For the circle \(C=x^2+y^2-6x+2y=0\),which of the following is incorrect:

• A. the radius of C is \( √10\)
• B. \((3,-1)\) lies inside of C
• C. \((7,3)\) lies inside of C
• D. the line\(the line x+3y=0 \)  intersects \(C\).
• E. one of diameters of \( C\) is not along \(x+3y=0\)

Answer 1

The Correct Option is D

Let's rewrite the equation of the circle in the standard form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. Given equation: $x^2 + y^2 - 6x + 2y = 0$ 

Complete the square for $x$ and $y$ terms: $(x^2 - 6x) + (y^2 + 2y) = 0$ $(x^2 - 6x + 9) + (y^2 + 2y + 1) = 0 + 9 + 1$ $(x-3)^2 + (y+1)^2 = 10$ So the center is $(3, -1)$ and the radius is $\sqrt{10}$. 

1. The radius of C is $\sqrt{10}$: Correct 

2. $(3,-1)$ lies inside of C: Substitute $(3,-1)$ into the equation: $(3-3)^2 + (-1+1)^2 = 0^2 + 0^2 = 0 < 10$. Since the distance from the center is 0, which is less than the radius $\sqrt{10}$, $(3, -1)$ lies inside the circle (it is the center). Correct 

3. $(7,3)$ lies inside of C: Substitute $(7,3)$ into the equation: $(7-3)^2 + (3+1)^2 = 4^2 + 4^2 = 16 + 16 = 32 > 10$. Since the distance from the center is greater than the radius, $(7,3)$ lies outside the circle. Incorrect 

4. The line $x+3y=0$ intersects C: We can write $x=-3y$. Substitute into the circle equation: $(-3y-3)^2 + (y+1)^2 = 10$ $9y^2 + 18y + 9 + y^2 + 2y + 1 = 10$ $10y^2 + 20y = 0$ $10y(y+2) = 0$ $y=0$ or $y=-2$ If $y=0$, $x=-3(0) = 0$. 
So, $(0,0)$ is an intersection point. If $y=-2$, $x=-3(-2) = 6$. So, $(6,-2)$ is an intersection point. 
Since there are intersection points, the line intersects the circle. Correct 

5. One of the diameters of C is not along $x+3y=0$: The equation $x+3y=0$ can be written as $y = -\frac{1}{3}x$. The line passes through the center $(3,-1)$ since $-1 = -\frac{1}{3}(3)$. Therefore, this line represents a diameter of the circle. Since there can be infinitely many diameters, one of them will not be along this line. Correct 

The incorrect statement is $(7,3)$ lies inside of C. Final Answer: 

The final answer is $\boxed{C}$

Answer 2

Step 1: Rewrite the circle equation in standard form.

The given equation of the circle is:

\[ x^2 + y^2 - 6x + 2y = 0 \]

To rewrite it in standard form, complete the square for both \( x \) and \( y \).

Step 2: Complete the square for \( x \).

For \( x^2 - 6x \):

\[ x^2 - 6x = (x - 3)^2 - 9 \]

Step 3: Complete the square for \( y \).

For \( y^2 + 2y \):

\[ y^2 + 2y = (y + 1)^2 - 1 \]

Step 4: Substitute back into the equation.

Substituting the completed squares into the original equation:

\[ (x - 3)^2 - 9 + (y + 1)^2 - 1 = 0 \]

Simplify:

\[ (x - 3)^2 + (y + 1)^2 = 10 \]

This is the standard form of the circle with:

• Center: \( (3, -1) \)
• Radius: \( \sqrt{10} \)

Step 5: Analyze each statement.

Statement 1: The radius of \( C \) is \( \sqrt{10} \).

This is correct, as derived above.

Statement 2: \( (3, -1) \) lies inside \( C \).

The center of the circle is \( (3, -1) \), which lies on the circle. This statement is Correct.

Statement 3: \( (7, 3) \) lies inside \( C \).

To check if \( (7, 3) \) lies inside the circle, calculate its distance from the center \( (3, -1) \):

\[ \text{Distance} = \sqrt{(7 - 3)^2 + (3 - (-1))^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} \]

Since \( \sqrt{32} > \sqrt{10} \), \( (7, 3) \) lies outside the circle. This statement is incorrect.

Statement 4: The line \( x + 3y = 0 \) intersects \( C \).

Substitute \( y = -\frac{x}{3} \) (from \( x + 3y = 0 \)) into the circle equation:

\[ x^2 + \left(-\frac{x}{3}\right)^2 - 6x + 2\left(-\frac{x}{3}\right) = 0 \]

Simplify:

\[ x^2 + \frac{x^2}{9} - 6x - \frac{2x}{3} = 0 \]

Multiply through by 9 to eliminate fractions:

\[ 9x^2 + x^2 - 54x - 6x = 0 \]

\[ 10x^2 - 60x = 0 \]

Factorize:

\[ 10x(x - 6) = 0 \]

\[ x = 0 \quad \text{or} \quad x = 6 \]

Thus, the line intersects the circle at two points. This statement is correct.

Statement 5: One of the diameters of \( C \) is not along \( x + 3y = 0 \).

The diameter of a circle passes through its center. The center of the circle is \( (3, -1) \). Substituting \( (3, -1) \) into \( x + 3y = 0 \):

\[ 3 + 3(-1) = 0 \]

This shows that the center lies on the line \( x + 3y = 0 \), so one of the diameters is along this line. This statement is Correct.

Final Answer:

The incorrect statement is: \( (7, 3) \) lies inside \( C \).