A tangent is drawn to the circle $2 x^{2}+2 y^{2}-3 x+4 y=0$ at point $A$ and it meets the line $x+y=3$ at $B(2,1)$, then $A B$ is equal to

• A

  $\sqrt{10}$
• Correct answer
  B

  2
• C

  $2 \sqrt{2}$
• D

  0

Solution

Equation of circle is, $\begin{array}{r} S \equiv 2 x^{2}+2 y^{2}-3 x+4 y=0 \\ \Rightarrow \quad S \equiv x^{2}+y^{2}-\frac{3}{2} x+2 y=0 \end{array}$

Here, $A B$ is the length of tangent to the circle from $B$ i.e., $\begin{aligned} A B &=\sqrt{(2)^{2}+(1)^{2}-\frac{3}{2}(2)+2(1)} \\ &=\sqrt{4+1-3+2}=\sqrt{4} \\ &=2 \text { units } \end{aligned}$

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