Question

If the equation of 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. $-7,32$

Answer 1

Answer: (A)

$5,-35$

According to question,

eq. of circle is $x^2+y^2-2x+6y-6=0$

The centre and radius of circle are $(1,-3)$ and $4$ respectively.

Since length of perpendicular from centre $(1,-3)$ to the tangent $3x-4y+k=0$ is equal to radius $4$.

Therefore,

=> $\left|\frac{3+12+k}{\sqrt{9+16}}\right|$= $4$

=>$15+k=20$ or $15+k=-20$

=> $k=5$ or $k=-35$