Question

The inverse of the point $(1,2)$ with respect to the circle $x^2+y^2-4x-6y+9=0$, is

• A. $\left(1,\frac{1}{2}\right)$
• B. $(2,1)$
• C. $(0,1)$
• D. $(1,0)$

Answer 1

The correct option is C

$(0,1)$

Explanation for correct option:

Equation of line:

Equation of $AO$ is

$$\begin{gathered} (y-2)=\frac{3-2}{2-1}(x-1) \\ \Rightarrow y=x+1 \end{gathered}$$

Let $P(h,h+1)$ is the inverse point of $(1,2)$ with the respect to the circle.

$$\begin{gathered} x^2+y^2-4x-6y+9=0 \\ \Rightarrow x^2-2\times 2\times x+{\left(2\right)}^2+y^2-2\times 3\times y+{\left(3\right)}^2={\left(2\right)}^2 \\ \Rightarrow {(x-2)}^2+{(y-3)}^2={\left(2\right)}^2 \end{gathered}$$

By comparing the equation with the general equation of circle ${(x-a)}^2+{(y-b)}^2=r^2$, where $(a,b)$ is the centre of the circle and $r$ is the radius of the cirlce.

So, center$=(1,2)$ and radius$=2$ .

Step-2 : Applying inverse point condition:

The points $QandP$ are inverse points with respect to the inversion circle if

$$\begin{gathered} PO\times QO=r^2 \\ \Rightarrow \sqrt{{(h-2)}^2+{(h+1-3)}^2}\times \sqrt{{(2-1)}^2+{(3-2)}^2}={\left(2\right)}^2 \\ \Rightarrow \sqrt{{(h-2)}^2+{(h-2)}^2}\times \sqrt{1+1}=4 \\ \Rightarrow \sqrt{2{(h-2)}^2}\times \sqrt{2}=4 \\ \Rightarrow \sqrt{{(h-2)}^2}\times 2=4 \\ \Rightarrow \pm (h-2)=2 \\ \Rightarrow h=4,h=0 \\ \Rightarrow h+1=5,h=1 \\ \Rightarrow (4,5),(0,1) \end{gathered}$$

Absurd $(4,5)$ as $P$ is on the same side and

Hence, $P(0,1)$ is the inverse of $(1,2)$.

Hence,option C is correct.