If one of the diameters of the curve $x^{2}+y^{2}-4 x-6 y+9=0$ is a chord of a circle with centre $(1,1),$ the radius of this circle is

• Correct answer
  A

  3
• B

  2
• C

  $\sqrt{2}$
• D

  1

Solution

Given circle $x^{2}+y^{2}-4 x-6 y+9=0$ $\therefore$ Centre $=\left(-\frac{1}{2} \times(-4),-\frac{1}{2} \times(-6)\right)$ $=(2.3)$ $\begin{aligned} \text { Radius } &=\sqrt{(-2)^{2}+(-3)^{2}-9} \\ &=\sqrt{4+9-9} \\ &=2 \end{aligned}$ $$ \begin{aligned} \therefore \quad A B &=\sqrt{(2-1)^{2}+(3-1)^{2}}=\sqrt{1+4}=\sqrt{5} \\ \therefore \quad A C &=\sqrt{A B^{2}+B C^{2}}=\sqrt{(\sqrt{5})^{2}+(2)^{2}} \\ &=\sqrt{5+4}=\sqrt{9}=3 \end{aligned} $$ Radius of required circle $=3$

Hence, the correct answer is option: 'A'.