Question

Consider three noncollinear points ABC with coordinates (0,6), (5,5), (-1,1) respectively equation of a line tangent to the circle circumscribing the triangle ABC and passing through the origin is

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

Answer 1

The correct option is D 2 X + 3 Y = 0

As shown in figure, circle is passing through three points $A(0,6)$, $B(5,5)$ and $C(-1,1)$

Now, general equation of circle is given by,

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

Point $A(0,6)$ lies on the circle. Thus, it must satisfy the equation of circle.

$\therefore {\left(0\right)}^2+{\left(6\right)}^2+2g\left(0\right)+2f\left(6\right)+c=0$

$\therefore 36+12f+c=0$

$\therefore c=-36-12f$ Equation (1)

Similarly, point $B(5,5)$ also lies on the circle. Thus, it must satisfy the equation of circle.

$\therefore {\left(5\right)}^2+{\left(5\right)}^2+2g\left(5\right)+2f\left(5\right)+c=0$

$\therefore 25+25+10g+10f+c=0$

$\therefore 50+10g+10f+c=0$

$\therefore c=-50-10g-10f$ Equation (2)

From equation (1) and (2), we get,

$-36-12f=-50-10g-10f$

Dividing both sides by -1, we get,

$36+12f=50+10g+10f$

$\therefore -10g+2f=14$ Equation (3)

Now, point $C(-1,1)$ also lies on the circle. Thus, it must satisfy the equation of circle.

$\therefore {(-1)}^2+{\left(1\right)}^2+2g(-1)+2f\left(1\right)+c=0$

$\therefore 1+1-2g+2f+c=0$

$\therefore 2-2g+2f+c=0$

$\therefore c=-2+2g-2f$ Equation (4)

From equation (1) and (4), we get,

$-36-12f=-2+2g-2f$

$\therefore 2g+10f+34=0$

Multiplying both sides by 5, we get,

$10g+50f=-170$ Equation (5)

Adding equations (3) and (5), we get,

$52f=-156$

$\therefore f=-3$

Put this value in equation (3), we get,

$-10g+2(-3)=14$

$\therefore -10g-6=14$

$\therefore -10g=20$

$\therefore g=-2$

Put these values in equation (4), we get,

$c=-2+2(-2)-2(-3)$

$c=-2-4+6$

$\therefore c=0$

Thus, equation of circle is given by,

$x^2+y^2+2(-2)x+2(-3)f+0=0$

$\therefore x^2+y^2-4x-6y=0$

Center of circle is $C(-g,-f)=C(2,3)$

Now, equation of tangent passing through point $(x_1,y_1)$ and tangential to the given circle is,

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

As our tangent is passing through origin, $(x_1,y_1)=(0,0)$

Thus, equation of tangent will become,

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

$-2x-3y=0$

Dividing both sides by -1, we get,

$2x+3y=0$

Thus, answer is option (D)