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
- A \(3 x-4 y-19=0,3 x-4 y+31=0\)
- B \(4 x+3 y-19=0,4 x+3 y+31=0\)
- C \(4 x+3 y+19=0,4 x+3 y-31=0\)
- D \(3 x-4 y+19=0,3 x-4 y+31=0\)
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 \(4 x+3 y-31=0\)
\(\Rightarrow \quad 6+k=\pm 25\)
\(\Rightarrow \quad k=19\) and \(-31\)
Hence, the tangent are \(4 x+3 y+19=0\) and \(4 x+3 y-31=0\)
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