Question

If a tangent drawn to the circle $x^{2}+y^{2}-6x-8y-11=0$ is perpendicular to the line $3x + 4y + k = 0$, then the distance from the origin to this tangent is

• A. 4
• B. 5
• C. 6
• D. 2

Hint:

The distance of a tangent line from the origin can be directly computed using $\frac{|c|}{\sqrt{a^2+b^2}}$ once the constant $c$ is determined via radius checking.

Answer 1

The Correct Option is C

Step 1: Find the circle centre and radius.
For $x^2+y^2-6x-8y-11=0$, the centre is $(3,4)$ and the radius is \[ R=\sqrt{3^2+4^2+11}=\sqrt{36}=6. \]
Step 2: Write the perpendicular family.
Any line perpendicular to $3x+4y+k=0$ looks like $4x-3y+c=0$ (we swap and change a sign).
Step 3: Apply the tangent condition.
A tangent touches the circle, so its distance from the centre equals the radius: \[ \frac{|4(3)-3(4)+c|}{\sqrt{4^2+3^2}}=6. \]
Step 4: Solve for $c$.
The top is $|0+c|$, so $\dfrac{|c|}{5}=6$, giving $|c|=30$, that is $c=\pm30$. The tangent is $4x-3y\pm30=0$.
Step 5: Distance from the origin.
\[ d=\frac{|4(0)-3(0)\pm30|}{\sqrt{4^2+3^2}}=\frac{30}{5}=6. \]
Step 6: State the result.
Both choices of sign give the same distance $6$. \[ \boxed{6} \]