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. $-7,32$

Answer 1

The correct option is A

$5,-35$

Explanation for the correct option

Step 1: Solve for the center and radius of the given circle

Given that, the equation of the tangent to the circle$x^2+y^2-2x+6y-6=0$ parallel to $3x-4y+7=0$ is $3x-4y+k=0$

The radius is perpendicular to the tangent at the point of contact.

Hence, the radius represents the perpendicular distance between the center of the circle and the tangent.

$x^2+y^2-2x+6y-6=0$ is the equation of the given circle.

Comparing the given equation to standard equation of circle $x^2+y^2+2gx+2fy+p=0$ we get

$g=-1,f=3,p=-6$

The center of the circle is given as $\left(-g,-f\right)$

Hence $(1,-3)=$$\left(x_1,y_1\right)$ is the center of the given circle

The radius of the circle is given as $r=\sqrt{g^2+f^2-p}$

$r=\sqrt{1+9+6}=\sqrt{16}=4$

Hence radius of the given circle is $4units$

Step 2: Solve for the required value

$3x-4y+k=0$ is the equation of tangent

Comparing given equation of tangent with $ax+by+c=0$ we get

$a=3,b=-4,c=k$

The perpendicular distance of a point $\left(x_1,y_1\right)$ from a line $ax+by+c=0$ is given as

$d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

Substituting the values we get

$\Rightarrow$ $4=\frac{\left|3\times 1-4\times \left(-3\right)+k\right|}{\sqrt{3^2+{\left(-4\right)}^2}}$

$\Rightarrow$ $20=\left|15+k\right|$

$\Rightarrow$$15+k=20$ or $-\left(15+k\right)=20$

$\Rightarrow$ $k=5$ or $k=-35$

Thus, the required values of $k$ are $5,-35$.

Hence, option (A) i.e. $5,-35$ is the correct answer.