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

Answer: (D)

the line$the line x+3y=0$ intersects $C$.

Answer 2

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.