Question

The center of the circle whose normals are $x^2-2xy-3x+6y=0$, is

• A. $\left(3,3/2\right)$
• B. $\left(3,-3/2\right)$
• C. $\left(3/2,3\right)$
• D. None of these

Answer 1

The correct option is A

$\left(3,3/2\right)$

Explanation for the correct option:

Finding the center of the circle:

The equation of normals is given $x^2-2xy-3x+6y=0$.

$$\begin{gathered} \Rightarrow x\left(x-2y\right)-3\left(x-2y\right)=0 \\ \Rightarrow \left(x-3\right)\left(x-2y\right)=0 \end{gathered}$$

So, the normals to the circle are line $x=3$ and $x=2y$.

The intersection point of these two normal is the center of the circle.

Therefore, substitute $x=3$ in $x=2y$.

$$\begin{gathered} \Rightarrow 3=2y \\ \Rightarrow y=3/2 \end{gathered}$$

Hence. center of the circle is $\left(3,\frac{3}{2}\right)$.

Therefore, the correct answer is option (A).