MHT CET · Maths · Circle
If \(m_{1}\) and \(m_{2}\) are the slopes of tangents to the circle \(x^{2}+y^{2}=4\) from the point \((3,2)\), then \(m_{1}-m_{2}\) is equal to
- A \(\frac{5}{12}\)
- B \(\frac{12}{5}\)
- C \(\frac{3}{2}\)
- D 0
Answer & Solution
Correct Answer
(B) \(\frac{12}{5}\)
Step-by-step Solution
Detailed explanation
Equation of pair of tangents is \(S S_{1}=T^{2}\)
\(\Rightarrow\left(x^{2}+y^{2}-4\right)(9+4-4)=(3 x+2 y-4)^{2}\)
\(\Rightarrow 5 y^{2}+16 y-12 x y+24 x-52=0\)
\(\therefore \quad m_{1}+m_{2}=\frac{-2 h}{b}=\frac{12}{5}\)
and \(\quad m_{1} m_{2}=0\)
Now, \(m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}\)
\(=\sqrt{\left(\frac{12}{5}\right)^{2}-0}\)
\(=\frac{12}{5}\)
\(\Rightarrow\left(x^{2}+y^{2}-4\right)(9+4-4)=(3 x+2 y-4)^{2}\)
\(\Rightarrow 5 y^{2}+16 y-12 x y+24 x-52=0\)
\(\therefore \quad m_{1}+m_{2}=\frac{-2 h}{b}=\frac{12}{5}\)
and \(\quad m_{1} m_{2}=0\)
Now, \(m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}\)
\(=\sqrt{\left(\frac{12}{5}\right)^{2}-0}\)
\(=\frac{12}{5}\)
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