TS EAMCET · Maths · Circle
If \(\theta\) is the angle between the tangents from \((-1,0)\) to the circle \(x^2+y^2-5 x+4 y-2=0\), then \(\theta\) is equal to
- A \(2 \tan ^{-1}\left(\frac{7}{4}\right)\)
- B \(\tan ^{-1}\left(\frac{7}{4}\right)\)
- C \(2 \cot ^{-1}\left(\frac{7}{4}\right)\)
- D \(\cot ^{-1}\left(\frac{7}{4}\right)\)
Answer & Solution
Correct Answer
(A) \(2 \tan ^{-1}\left(\frac{7}{4}\right)\)
Step-by-step Solution
Detailed explanation
We know that, the angle between the two tangents from \((\alpha, \beta)\) to the circle \(x^2+y^2=r^2\) is \[ 2 \tan ^{-1} \frac{r}{\sqrt{S_1}} \] Given equation of circle is \[ x^2+y^2-5 x+4 y-2=0 \text {. } \] Now, radius,…
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