KCET · Maths · Circle
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 \(\sqrt{10}\)
- B 2
- C \(2 \sqrt{2}\)
- D 0
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
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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