Question

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is ${45}^o$, find $k$.

Answer 1

As we know that angle between the two circles is given as-

$\cos \theta =\frac{{r_1}^2+{r_2}^2-d^2}{r_1{r}_2}$

Where,

$r_1=$ Radius of first ciircle

$r_2=$ Radius of second circle

$d=$ Distance between the centre of two circles

Equation of first circles-

$x^2+y^2-12x-6y+41=0$

${(x-6)}^2+{(y-3)}^2-36-9+41=0$

${(x-6)}^2+{(y-3)}^2={\left(2\right)}^2$

Here,

$r_1=2$

$C_1=(6,3)$

Equation of another circle-

$x^2+y^2+kx+6y-59=0$

${(x+\frac{k}{2})}^2+{(y+3)}^2-{\left(\frac{k}{2}\right)}^2-9-59=0$

${(x+\frac{k}{2})}^2+{(y+3)}^2=68+\frac{k^2}{4}$

Here,

$r_2=\sqrt{68+\frac{k^2}{4}}$

$C_2=(-\frac{k}{2},-3)$

Now,

$C_1{C}_2=\sqrt{{(-\frac{k}{2}-6)}^2+{(-3-3)}^2}=\sqrt{\frac{k^2}{4}+6k+72}$

$\theta =45°\left(\text{Given}\right)$

Therefore,

$\cos 45°=\frac{4+(68+\frac{k^2}{4})-(\frac{k^2}{4}+6k+72)}{2\left(\sqrt{68-\frac{k^2}{4}}\right)}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{6k}{\sqrt{68-\frac{k^2}{4}}}$

$\Rightarrow \frac{\sqrt{68-\frac{k^2}{4}}}{6k}=\sqrt{2}$

Squaring both sides, we have

$\Rightarrow {\left(\frac{\sqrt{68-\frac{k^2}{4}}}{6k}\right)}^2={\left(\sqrt{2}\right)}^2$

$\frac{68-\frac{k^2}{4}}{36k^2}=2$

$\Rightarrow 68-\frac{k^2}{4}=72k^2$

$\Rightarrow \frac{289k^2}{4}=68$

$\Rightarrow k=\pm \sqrt{\frac{272}{289}}=\pm \frac{4}{\sqrt{17}}$

Hence the value of $k$ is $\pm \frac{4}{\sqrt{17}}$.