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MHT CET · Maths · Circle

The equations of the tangents to the circle \(x^{2}+y^{2}-6 x+4 y=12\), which are parallel to the straight line \(4 x+3 y+5=0\), are

  1. A \(3 x-4 y-19=0,3 x-4 y+31=0\)
  2. B \(4 x+3 y-19=0,4 x+3 y+31=0\)
  3. C \(4 x+3 y+19=0,4 x+3 y-31=0\)
  4. D \(3 x-4 y+19=0,3 x-4 y+31=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4 x+3 y+19=0,4 x+3 y-31=0\)

Step-by-step Solution

Detailed explanation

Let equation of tangent be \(4 x+3 y+k=0\) centre of the circle is \((3,-2)\). Then, \(\quad \sqrt{9+4+12}=\left|\frac{4(3)+3(-2)+k}{\sqrt{16+9}}\right|\)
\(\Rightarrow \quad 6+k=\pm 25\)
\(\Rightarrow \quad k=19\) and \(-31\)
Hence, the tangent are \(4 x+3 y+19=0\) and \(4 x+3 y-31=0\)