Question

Find the equation of the tangent at P of the circle S=0 where P and S are given by i) P=(−6,−9)Sx2 +y2 +4x+6y−39

Answer 1

$\textbf{Concept used:}$

$\text{The equation of tangent at the point (x_1,y_1) }$

$\text{to the circle x^2+y^2+2gx+2fy+c=0 is}$

$\boxed{\bf\,xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0}$

$\textbf{Given:}$

$\text{Equation of the given circle: x^2+y^2+4x+6y-39=0 }$

$\text{Here, g=2, f=3 and c=-39}$

$\text{The equation of the tangent at P(-6,-9) is}$

$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$

$x(-6)+y(-9)+2(x-6)+3(y-9)-39=0$

$-6x-9y+2x-12+3y-27-39=0$

$-4x-6y-78=0$

$\implies\boxed{\bf\,2x+3y-39=0}$

$\therefore\textbf{The equation of tangent P is 2x+3y-39=0}$

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