Question

The length (in units) of the tangent from point $\left(5,1\right)$ to the circle $x^2+y^2+6x-4y-3=0$ is

• A. $81$
• B. $29$
• C. $7$
• D. $21$

Answer 1

The correct option is C

$7$

Explanation for the correct option:

Step 1: State given data and formula used

The given equation of the circle is $x^2+y^2+6x-4y-3=0$.

And, the coordinates of the point from which the tangent is drawn are $\left(5,1\right)$.

Now, we know that if a circle is represented by the equation $x^2+y^2+2gx+2fy+c=0$, then the length of the tangent drawn to the circle from an external point $P\left(x_1,y_1\right)$ can be calculated as,

Length of tangent $=\sqrt{x_1^2+y_1^2+2g{x}_1+2f{y}_1+c}$

Step 2: Calculate the length of the tangent

According to the question, a tangent to the circle $x^2+y^2+6x-4y-3=0$ is drawn from the point $\left(5,1\right)$.

So, $\left(x_1,y_1\right)=\left(5,1\right)$

Then, the length of the tangent $=\sqrt{x_1^2+y_1^2+6x_1-4y_1-3}$

$\Rightarrow$ $=\sqrt{{\left(5\right)}^2+{\left(1\right)}^2+6\left(5\right)-4\left(1\right)-3}$

$\Rightarrow$ $=\sqrt{25+1+30-4-3}$

$\Rightarrow$ $=\sqrt{49}$

$\Rightarrow$ $=7\mathrm{units}$

Hence, option C is the correct option.