Question

If the equation of tangent to the circle $x^2+y^2-2x+6y-6=0$ and 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

$5,-35$

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

Given equation of circle is $x^2+y^2-2x+6y-6=0$

For the general equation of the circle is $x^2+y^2+2gx+2fy+c=0$ the radius is $\sqrt{g^2+f^2-c}$ and the center is $\left(-g,-f\right)$

Now comparing with the general equation of the circle we have

$g=-1,f=3$ and $c=-6$

Therefore radius of the circle is $=\sqrt{g^2+f^2-c}$

$$\begin{gathered} =\sqrt{{\left(-1\right)}^2+{\left(3\right)}^2-\left(-6\right)} \\ =\sqrt{1+9+6} \\ =4 \end{gathered}$$

The center is $\left(-g,-f\right)=\left(1,-3\right)$

Step 2: Solve for the value of $k$

We know that the distance of the point $\left(x_1,y_1\right)$ from the straight line $ax+by+c=0$ is $\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

Therefore the distance of the point $\left(1,-3\right)$ from the tangent $3x-4y+k=0$ is the radius of the circle

$$\begin{gathered} \therefore \frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}=4 \\ \Rightarrow \frac{\left|3\left(1\right)-4\left(-3\right)+k\right|}{\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}}=4 \\ \Rightarrow \frac{\left|15+k\right|}{5}=4 \\ \Rightarrow \left|15+k\right|=20 \\ \Rightarrow 15+k=20\mathrm{or}15+\mathrm{k}=-20 \\ \Rightarrow \mathrm{k}=5\mathrm{or}\mathrm{k}=-35 \end{gathered}$$

Hence, Option (A) is correct i.e. $5,-35$