Question

If $x^2+y^2-5x-14y-34=0,x^2+y^2+2x+4y+k=0$ circles are orthogonal then find ${}^{\prime }{k}^{\prime }.$

Answer 1

Equation of circle-

$S_1:x^2+y^2-5x-14y-34=0\Rightarrow {(x-\frac{5}{2})}^2+{(y-7)}^2-\frac{25}{4}-49-34=0$

$\Rightarrow S_1:{(x-\frac{5}{2})}^2+{(y-7)}^2={\left(\frac{\sqrt{357}}{2}\right)}^2$

Here $r_1=\frac{\sqrt{357}}{2}$

$C_1=(\frac{5}{2},7)$

$S_2{x}^2+y^2+2x+4y+k=0\Rightarrow {(x+1)}^2+{(y+2)}^2-1-4+k=0$

$\Rightarrow S_1:{(x+1)}^2+{(y+2)}^2={\left(\sqrt{5-k}\right)}^2$

Here $r_1=\sqrt{5-k}$

$C_1=(-1,-2)$

: $C_1{C}_2=\sqrt{{(-1-\frac{5}{2})}^2+{(-2-7)}^2}=\frac{\sqrt{373}}{2}$

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

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

Here $d$ is the distance between centre of circle.

Since the circles orthogonal, i.e., $\theta =\frac{\pi }{2}$

$\therefore \cos \frac{\pi }{2}=\frac{{\left(\frac{\sqrt{357}}{2}\right)}^2+{\left(\sqrt{5-k}\right)}^2-{\left(\frac{\sqrt{373}}{2}\right)}^2}{r_1{r}_2}$

$\frac{357}{4}+5-k-\frac{373}{4}=0$

$357+20-4k-373=0$

$k=0$

Hence the value of $k$ for which the circles are orthogonal is $0$.