AP EAMCET · Maths · Circle
If \(2 x-3 y+3=0\) and \(x+2 y+k=0\) are conjugate lines with respect to the circle \(\mathrm{S} \equiv x^2+y^2+8 x-6 y-24=0\), then the length of the tangent drawn from the point \(\left(\frac{k}{4}, \frac{k}{3}\right)\) to the circle \(\mathrm{S}=0\) is
- A \(7\)
- B \(1\)
- C \(12\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}& \text { Given equation of circle } x^2+y^2+8 x-6 y-24=0 \\& \Rightarrow(x+4)^2+(y-3)^2=24+25=49\end{aligned}\) So, centre \(=(-4,3)\), radius \(=7\) Since, \(2 x-3 y+3=0\) and \(x+2 y+k=0\) are conjugate lines with respect to the given circle. So,…
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