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

Given that:

\( C: x^2 + y^2 - 6x + 2y = 0,\)

 we must analyze the given options:

first check option 2

Option 2: The center of circle C is \((3, -1).\)

To find the center of the circle, we need to complete the square for the \(x\) and \(y\) terms. 

Rewrite the equation as:

\((x^2 - 6x) + (y^2 + 2y) = 0\)

 for \(x^2 - 6x\), we add  \(\dfrac{6}{2}^{2} = 9 \) to make complete the square.

\((x^2 - 6x + 9) + (y^2 + 2y) = 9\)

Similarly, for \(y^2 + 2y\), we add and subtract \((2/2)^2 = 1\):

\((x^2 - 6x + 9) + (y^2 + 2y + 1) = 9 + 1\)

⇒\((x - 3)^2 + (y + 1)^2 = 10\)

Comparing the above expression with the ,Circle in the standard form i.e.\( (x - h)^2 + (y - k)^2 = r^2\) , where \((h, k)\) is the center of the circle.

 center of circle C is \((h, k) = (3, -1)\). So option 1 is correct.

Option 1: The radius of circle C is \(√10\).

Comparing with the standard form of equation we get  this option is  correct also.

Option 3: Solving in similar manner we get this option also stands correct.

Option 4 :  The line \(x+3y=0\) does not intersect with the circle equation as the  real values of x and y  does not satisfy the circle equation here.

Hence automatically the option 5  is correct .

So we can now state that as the question is asking about the incorrect option so the option 4 is incorrect  and is the desired answer.