Question

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

• A. $ 3x - 4y -1 9 = 0 $ , $ 3x - 4y + 31 = 0 $
• B. $ 4x + 3y - 19 = 0 $ , $ 4x + 3y + 31 = 0 $
• C. $ 4x + 3y + 19 = 0 $ , $ 4x + 3y - 31 = 0 $
• D. $ 3x - 4y + 19 = 0 $ , $ 3x - 4y + 31 = 0 $

Answer 1

Answer: (C)

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