Question

Consider the circle $x^2+y^2-6x+4y=12$. The equation of a tangent to this circle that is parallel to the line $4x+3y+5=0$ is

• A. 4 x + 3 y + 10 = 0
• B. 4 x + 3 y - 9 = 0
• C. 4 x + 3 y + 9 = 0
• D. 4 x + 3 y - 31 = 0

Answer 1

Answer: (D)

We have,
$x^2+y^2-6x+4y=12$
$\Rightarrow (x-3{)}^2+(y+2{)}^2=25$
$\Rightarrow x-3{)}^2+(y+2{)}^2=(5{)}^2$
Equation of tangent whose slope $m$ is
$y+2=m(x-3)\pm 5\sqrt{m^2+1}.....(i)$
Now, this tangent is parallel to line $4x+3y+5=0$
$\therefore$ Slope of line is $-\frac{4}{3}$
Put the value of $m=-\frac{4}{3}$ is Eq. (i), we get
$y+2=-\frac{4}{3}(x-3)\pm 5\sqrt{{\left(\frac{-4}{3}\right)}^2+1}$
$\Rightarrow y+2=\frac{-4}{3}(x-3)\pm 5\left(\frac{5}{3}\right)$
$\Rightarrow 3y+6=-4x+12\pm 25$
$\Rightarrow 4x+3y=6\pm 25$
$\Rightarrow 4x+3y=31$ or $4x+3y=-19$
Hence, equation of tangent is
$4x+3y-31=0$ or $4x+3y+19=0$