Question

Equations of tangents perpendicular to the line $2x+3y+7=0$ to the circle $x^2+y^2-2x+4y+1=0$ are :

• A. $3x-2y-7\pm 2\sqrt{13}=0$
• B. $3x-2y\pm 2\sqrt{13}=0$
• C. $3x-2y+7\pm 2\sqrt{13}=0$
• D. None of these

Answer 1

Answer: (A)

$3x-2y-7\pm 2\sqrt{13}=0$

Slope of tangent $=\frac{3}{2}$

Centre of $x^2+y^2-2x+4y+1=0$ is $(1,-2)$

Radius $=\sqrt{1^2+2^2-1}=2$

tangent be $y=\frac{3}{2}x+C$

$2y-3x-2c=0$

Distance from centre to this line = radius

$\left|\frac{-4-3-2C}{\sqrt{13}}\right|=2$

$|-7-2C|=2\sqrt{13}$

$2C+7=\pm 2\sqrt{13}$

$2C=-7\pm 2\sqrt{13}$

Tangent $\to 2y-3x+7\pm 2\sqrt{13}=0$

$3x-2y-7\pm 2\sqrt{13}=0$

$A$ is Correct.