Question

If the equation of the tangent to the circle $x^2+y^2-2x+6y-6=0$ parallel to $3x-4y+7=0$ is $3x-4y+k=0,$ then the values of $k$ are:

• A. $5,-35$
• B. $-5,35$
• C. $7,-32$
• D. $-7,32$
• E. $3,-13$

Answer 1

Answer: (A)

The given equation of circle is $x^2+y^2-2x+6y-6=0$ The centre and radius of circle are $(1,-3)$ and 4 respectively. Length of perpendicular from $(1,-3)$ to $3x-4y+k=0$ is equal to radius 4. $\therefore$ $\left|\frac{3+12+k}{\sqrt{9+16}}\right|=4$ $\Rightarrow$ $15+k=\pm 20$ $\Rightarrow$ $k=20-15=5$ Or $k=-20-15=-35$