Question

If $3x+y+k=0$ is a tangent to the circle $x^2+y^2=10$ the values of k are,

• A. $\pm 5$
• B. $\pm 7$
• C. $\pm 9$
• D. $\pm 10$

Answer 1

Answer: (D)

Given, line is $3x+y+k=0$
$\Rightarrow y=-3x-k$
And equation of circle is $x^2+y^2=10$
Here, $a^2=10,m=-3,c=-k$
If given line touches the circle , then length of intercept $=0$
$\Rightarrow 2\sqrt{\frac{a^2\left(1+m^2\right)-c^2}{1+m^2}}=0$
$\Rightarrow 2\sqrt{\frac{10\left(1+9\right)-k^2}{1+9}}=0$
$\Rightarrow \sqrt{100-k^2}=0$
$\Rightarrow 100-k^2=0$
$\Rightarrow k=\pm 10$
Alternative : If the given line is tangent to the circle, then the length of the perpendicular from the centre upon the line is equal to the radius of the circle.
ie, $\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|=r$
$\Rightarrow \left|\frac{3\times 0+6\times 0+k}{\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}}\right|=\sqrt{10}$
$\Rightarrow \left|\frac{k}{\sqrt{10}}\right|=\sqrt{10}$
$\Rightarrow k=\sqrt{100}$
$\Rightarrow k=\pm 10$