Question

If $S=0$ is a circle and $P(x_1,y_1)$ is an exterior point with respect to $S=0$ then the length of the tangent from $P(x_1,y_1)$ to $S=0$ is $\sqrt{S_{11}}$.

Answer 1

Let $S=0$ i.e. $x^2+y^2+2gx+2gx+2fy+c=0$ be the equation of circle $P\left(x_1{y}_1\right)$ be the external point and PT & PS are 2 tangents to give circle.

Since, $\Delta PTC$ is a right angled triangle.

$\Rightarrow P{T}^2+T{C}^2=P{C}^2$ __(A)

But $TC$ = radius of circle $=\sqrt{g^2+f^2-c}$

$\Rightarrow T{C}^2=g^2+f^2-c$ __(I)

Also, $P{C}^2=[x_1-(g){]}^2+[y_1-(-f){]}^2$

$P{C}^2=(x_1+g{)}^2+(y_1+f{)}^2$ __(II)

Using (I) & (II) in equation A, we get

$P{T}^2+g^2+f^2-C=(x_1+g{)}^2+(y_1+f{)}^2$

$\Rightarrow P{T}^2+g^2+f^2-c=x_1^2+g^2+2g{x}_1+g^2+y_1^2+2f{y}_1+f^2$

$\Rightarrow P{T}^2=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c$

$\Rightarrow |PT|=\sqrt{x_1^2+y_1^2=2g{x}_1+2f{y}_1+c}=\sqrt{s_{11}}$ Hence proved