Question

The equation of tangent to the circle $x^2+y^2+2gx+2fy+c=0$ at its point $(x_1,y_1)$ is given by $x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c=0$

• A. True
• B. False

Answer 1

The correct option is A

True

If we know the slope of the tangent , we can find the equation of tangent at $(x_1,y_1)$

Tangent and the radius CP are perpendicular to each other.

$\Rightarrow Slopeoftangent\times slopeofCP=-1$

$\Rightarrow slopeoftangent=\frac{-1}{slopeofCP}$

Slope of CP $=\frac{y_1+f}{x_1+g}$

$\Rightarrow slopeoftangent=\frac{(x_1+g)}{y_1+f}$

We can now write the equation of tangent $(y-y_1)=\frac{x_1+g}{y_1+f}(x-x_1)$

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

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

$\Rightarrow y.x{x}_1+y{y}_1+fy-f{y}_1+xg-x_{1}g-x_1^2-y_1^2=0$

$\Rightarrow x{x}_1+y{y}_1+xg+fy=x_1^2-y_1^{2}f{g}_1+x_{1}g$ ................(1)

$\Rightarrow x_1^2-y_1^2+2g{x}_1+2f{g}_1+c=0$

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

(1),(2) $\Rightarrow x{x}_1+y{y}_1+xg+fy=-g{x}_1-f{y}_1-c$

$\Rightarrow x{x}_1+y{y}_1+(x+x_1)g+(y+y_1)f+c=0$

$\Rightarrow$ the given statement is correct.