Question

Equation of the pair of tangents drawn from the origin to the circle $x^2+y^2+2gx+2fy+c=0$ is

• A. $gx+fy+c\left(x^2+y^2\right)$
• B. ${\left(gx+fy\right)}^2=x^2+y^2$
• C. ${\left(gx+fy\right)}^2=c^2\left(x^2+y^2\right)$
• D. ${\left(gx+fy\right)}^2=c\left(x^2+y^2\right)$

Answer 1

The correct option is D ${\left(gx+fy\right)}^2=c\left(x^2+y^2\right)$

The equation of pair of tangent to the circle $x^2+y^2+2gx+2fy+c=0$ from the point $(x_1,y_1)$ is

$(x^2+y^2+2gx+2fy+c)({x_1}^2+{y_1}^2+2g{x}_1+2f{y}_1+c)={[x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c]}^2$

Thus, the equation of pair of the tangent to the circle $x^2+y^2+2gx+2fy+c=0$ from the origin is

$(x^2+y^2+2gx+2fy+c)(0+0+2g\times 0+2f\times 0+c)={[x\times 0+y\times 0+g(x+\times 0)+f(y+\times 0)]}^2$

$\Rightarrow c(x^2+y^2+2gx+2fy+c)={(gx+fy+c)}^2$

$\Rightarrow c{x}^2+c{y}^2+2cgx+2cfy+c^2=g^2{x}^2+f^2{y}^2+c^2+2gfxy+2fyc+2gcx$

$\Rightarrow c{x}^2+c{y}^2={\left(gx\right)}^2+{\left(fy\right)}^2+2gfxy$

$\Rightarrow c(x^2+y^2)={(gx+fy)}^2$

or ${(gx+fy)}^2=c(x^2+y^2)$