MHT CET · Maths · Circle
The equations of the tangents to the circle \(x^2+y^2=36\) which are perpendicular to the line \(5 x+y=2\), are
- A \(x+5 y \pm 6 \sqrt{26}=0\)
- B \(x-5 y \pm 6 \sqrt{26}=0\)
- C \(5 x-y \pm 6 \sqrt{26}=0\)
- D \(5 x+y \pm 6 \sqrt{26}=0\)
Answer & Solution
Correct Answer
(B) \(x-5 y \pm 6 \sqrt{26}=0\)
Step-by-step Solution
Detailed explanation
\(r^2 = 36 \Rightarrow r=6\) Slope of \(5x+y=2\) is \(m_1 = -5\).
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