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

The Correct Option is 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, $\left|\frac{ax_{1} + by_{1} +c}{\sqrt{a^{2} +b^{2}}}\right| = r$
$ \Rightarrow \left|\frac{3 \times0 + 6 \times0+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 $