Question

Prove:
If two tangents PT1 & PT2 are drawn from the point P(x1, y1) to the circle S = x2 + y2 + 2gx + 2fy + c = 0,
then the equation of the chord of contact T1T2 is: xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0.

Answer 1

Dear Student,

The equation of the circle is $x^2+y^2+2gx+2fy+c=0$.
Let the tangent $P{T}_1$ drawn from the point $P\left(x_1,y_1\right)$ touch the circle at $T_1\left(a,b\right)$ and the tangent $P{T}_2$drawn from the point $P\left(x_1,y_1\right)$ touch the circle at $T_2\left(d,e\right)$.
Then the equation of the tangent $P{T}_1$ is given as $ax+by+g\left(x+a\right)+f\left(y+b\right)+c=0$.
The the equation of the tangent $P{T}_2$ is given as $dx+ey+g\left(x+d\right)+f\left(y+e\right)+c=0$.
The point $P\left(x_1,y_1\right)$ satisfies both the equation of the tangent. So,
$a{x}_1+b{y}_1+g\left(x_1+a\right)+f\left(y_1+b\right)+c=0.....\left(i\right)$
$d{x}_1+e{y}_1+g\left(x_1+d\right)+f\left(y_1+e\right)+c=0.....\left(ii\right)$
The equation (i) and (ii) shows that the points (a, b) and (c, d) lies on the line $x{x}_1+y{y}_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$.
Thus, the straight line ​$x{x}_1+y{y}_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$ represents the chord of contact $T_1{T}_2$.
Regards,