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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
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The correct answer is:
2
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.,
$\begin{aligned}
A B &=\sqrt{(2)^{2}+(1)^{2}-\frac{3}{2}(2)+2(1)} \\
&=\sqrt{4+1-3+2}=\sqrt{4} \\
&=2 \text { units }
\end{aligned}$
$\begin{array}{r}
S \equiv 2 x^{2}+2 y^{2}-3 x+4 y=0 \\
\Rightarrow \quad S \equiv x^{2}+y^{2}-\frac{3}{2} x+2 y=0
\end{array}$
Here, $A B$ is the length of tangent to the circle from $B$ i.e.,
$\begin{aligned}
A B &=\sqrt{(2)^{2}+(1)^{2}-\frac{3}{2}(2)+2(1)} \\
&=\sqrt{4+1-3+2}=\sqrt{4} \\
&=2 \text { units }
\end{aligned}$
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