Question

Consider two circles passing through two distinct points $(0,a)$ and $(0,-a)$ on $y$-axis and touching the line $y=3x+4$. If the circles are orthogonal and $a=\pm \frac{k_1}{\sqrt{k_2}}$, where $k_1,k_2$ are co-prime, then the value of $k_1+k_2$ is

Answer 1

Let the equation of circle be
$x^2+y^2+2gx+2fy+c=0\cdots \left(1\right)$
Putting the points $(0,a)$ and $(0,-a)$ simultaneously,
$a^2+2fa+c=0\cdots \left(2\right)a^2-2fa+c=0\cdots \left(3\right)$
As both the points are distinct, so, $a\ne 0$
Using equation $\left(2\right)$ and $\left(3\right)$, we get
$c=-a^2,f=0$
From $\left(1\right)$, we get
$x^2+y^2+2gx-a^2=0$
Centre and radius of the circle is,
$C(-g,0),r=\sqrt{g^2+a^2}$

Now, the given tangent is,
$y=3x+4\Rightarrow 3x-y+4=0$
Distance from the centre to the tangent $=$ Radius of the circle
$\left|\frac{-3g+4}{\sqrt{10}}\right|=\sqrt{g^2+a^2}$
$\Rightarrow (3g-4{)}^2=10(g^2+a^2)\Rightarrow g^2+24g+(10a^2-16)=0$

Let the two roots of the equation be $g_1,g_2$.
$\therefore g_1{g}_2=10a^2-16$
Now, the two circles are
$x^2+y^2+2g_{1}x-a^2=0x^2+y^2+2g_{2}x-a^2=0$
The circles are orthogonal if,
$2g_1{g}_2+0=-a^2-a^2\Rightarrow g_1{g}_2=-a^2$
$\Rightarrow 10a^2-16=-a^2\Rightarrow a=\pm \frac{4}{\sqrt{11}}\therefore k_1+k_2=15$