Question

Consider $3$ non collinear points $A,B,C$ with coordinates $(0,6),(5,5)$ and $(-1,1)$ respectively. Equation of a line tangent to the circle circumscribing the triangles $ABC$ and passin through the origin is

• A. $2x-3y=0$
• B. $3x+2y=0$
• C. $3x-2y=0$
• D. $2x+3y=0$

Answer 1

The correct option is A $2x-3y=0$

contd equation of the circle by putting values of g=-2,f=-3,c=0

$x^2+y^2+2gx+2fy+c=0$$

$=>x^2+y^2+2\ast (-2)\ast x+2\ast (-3)\ast y+0=0$

$=>x^2+y^2-4x-6y=0$

So since here the tangent of the circle passes through the origin:-

$xx1+yy1-2(x+x1)-3(y+y1)=0$

$=>0+0-2(x+0)-3(y+0)=0$

$=>-2x-3y=0$

$=.2x+3y=0\left(ans\right)$