Find the equation of circle (s) touching the $x$-axis at a distance of 3 units from the origin and having an intercept of length $2\sqrt{7}$ on the Y-axis?

Let $\quad x^2+y^2+2 y x+2 y+c=0$ the only point-which lies both the circle and the $x$-axis is $(3,0)$

$(a,0)$

$a^2+2 g a+c=0$

$\Rightarrow \quad g^2=c$

$9+6 g+c=0$

$9+6 g+g^2=0 \Rightarrow(g+3)^2=0$
$\Rightarrow c = 9 ,g=-3$

$\begin{gathered}2 \sqrt{f^2-c}=2 \sqrt{f^2-9}=2\sqrt{7} \\ f= \pm 4 .\end{gathered}$

$C_1: \$ $\ g=-3, f=-4$

Centre $(-g , -f) = (3 , 4)$

$C_2: g=-3, f=+4$

Centre $(-g , -f) = (3 , -4 )$

Let $L_1$, be a straight line passing through the origin and $L_2$ be the straight line $x+y=1$. If the intercepts made by the circle $x^2+y^2-x+3 y=0$ on $L_1$ and $L_2$ are equal, then which of the following equation can represent $L_1$ ?

a) $x+y=0$

b) $x-y=0$

c) $x+7 y=0$

d) $x-7 y=0$

$\begin{aligned} C: \quad x^2+y^2-x & +3 y=0 \\ L_2: \quad x+y & =1 \\ y & =1-x\end{aligned}$

$x^2+(1-x)^2-x+3(1-x)=0$

$\Rightarrow \quad x^2-3 x+2=0$

$\Rightarrow \quad x=1,2$

$x^2+m^2 x^2-x+3 m x=0$

$x^2\left(1+m^2\right)-(1-3 m) x=0$

$x=0,(1-3 m) /\left(1+m^2\right)$

$\therefore L_1$ intersects the circle at- $(0,0)$ $(y=mx)$ and $\left(\frac{1-3 m}{1+m^2}, \frac{m(1-3 m)}{1+m^2}\right)$

$\therefore \sqrt{2}=\sqrt{\frac{(1-3 m)^2}{\left(1+m^2\right)}}$

$\Rightarrow \quad 2=(-3 m)^2 /\left(1+m^2\right)$

$\Rightarrow \quad 2+2 m^2=1-6 m+9 m^2$

$\Rightarrow \quad 7 m^2-6 m-1=0$

$\Rightarrow \quad(m-1)(7 m+1)=0$

$\Rightarrow m=1,-\frac{1}{7}$.

$\therefore L_1, y=x$ or $\ y=\frac{-x}{7}$.

$C: x^2+y^2+2 g x+2 f y+c=0$
$P\left(x_1, y_1\right)$

Slope of tangent = $\frac{(y-y_1)}{(x-x_1)}$

$\frac{y_1-(-f)}{x_1-(-g)}=\frac{y_1+f}{x_1+g}=\text { slope of OP}$

$\frac{\left(y-y_1\right)}{\left(x-x_1\right)} \times \frac{\left(y_1+f\right)}{\left(x_1+y\right)}=-1$

$\Rightarrow \ \left(y-y_1\right)\left(y_1+f\right)+\left(x-x_1\right)\left(x_1+g\right)=0$

$\Rightarrow y y_1 +x x_1 +y f+x g= y_1^2 + y_1 f+ x_1^2 + x_1 g$

since $P\left(x_1, y_1\right)$ lies on circle $C$,

$x_1^2+y_1^2+2 g x_1+2 f y_1+c=0 \\ x_1^2+y_1^2+g x_1+f y_1={-\left(g x_1+f y_1\right.}+c)$

$y y_1+x x_1+y f+x g+g x_1+f y_1+c=0$

$y\left(y_1+f\right)+x\left(x_1+g\right)+g x_1+f y_1+c=0$

$C: x^2+y^2+2 g x+2 fy+c=0$

$\text { slope } of \text { the normal at } P$

$=\left(y_2-(-f)\right) /\left(x_2-(-g)\right)$

$=(y-(-f)) /(x-(-g))$

$\frac{(\left.y_2+f\right)}{\left(x_2+g\right)}=\frac{(y+f)}{(x+g)}$

$\Rightarrow \ x\left(y_2+f\right)+g\left(y_2+f\right)=y\left(x_2+g\right)+f\left(x_2+g\right)$

$\text { i.e., } \ x\left(y_2+f\right)-y\left(x_2+g\right)$

$=f\left(x_2+g\right) -g(y_2+f)$

$=\left(f x_2-g y_2\right)$

$C: x^2+y^2+2 g x+2 f y+c=0$

$O P= \sqrt{\left(x_1-(-g)\right)^2+\left(y_1-(-f)\right)^2}$

$O P=\sqrt{\left(x_1+g\right)^2+\left(y_1+f\right)^2}$

$\because O P^2=O T^2+P T^2$

$\text { i.e., }$

$P T=\sqrt{O P^2-O T^2}$

$P T=\sqrt{\left(x_1+g\right)^2+\left(y_1+f\right)^2-\left(g^2+f^2-c\right)}$