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 - 19 = 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

The correct option is C 4x + 3y + 19 = 0, 4x + 3y - 31 = 0

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