Question

If the tangents be drawn to the circle $x^2+y^2=12$ at its points of intersection with the circle $x^2+y^2-5x+3y-2=0$, then the tangents intersect at the point

• A. $(6,\frac{18}{5})$
• B. $(-6,-\frac{18}{5})$
• C. $(6,-\frac{18}{5})$
• D. $(-6,\frac{18}{5})$

Answer 1

The correct option is C $(6,-\frac{18}{5})$

$x^2+y^2-5x+3y-2=0x^2+y^2=12$
Chord of contact through $(a,b)$ ax+by+g(x+a)+f(y+b)+c=0 c:\quad ax+by-12=0$Commonchordequation$=S'-S -5x+3y-2+12=0\\ -5x+3y+10=0\\ -\cfrac { a }{ 5 } =\cfrac { b }{ 3 } =\cfrac { -12 }{ 10 } \\ -\cfrac { a }{ 5 } =\cfrac { -12 }{ 10 } \quad \quad \quad \quad \cfrac { b }{ 3 } =\cfrac { -12 }{ 10 } \\ a=6\quad \quad \quad \quad \quad \quad \quad b=\cfrac { -18 }{ 5 } $Theyintersectat$ (6,-\cfrac { 18 }{ 5 } )$