Question

The equation of the pair of tangents drown to the circle $x^2+y^2-2x+4y+3=0$ from (6. -5) is-

• A. ${7x}^2+23y^2+30xy+66x+50y-73=0$
• B. ${7x}^2+23y^2+30xy-66x-50y+73=0$
• C. ${7x}^2+23y^2+30xy-66x-50y-73=0$
• D. None of these

Answer 1

The correct option is D None of these

Equation of circle is given by,

$S:x^2+y^2-2x+4y+3=0$

Compare this equation with standard form of equation i.e. $x^2+y^2+2gx+2fy+c=0$, we get,

$2g=-2$

$\therefore g=-1$

$2f=4$

$\therefore f=2$

$c=3$

Now, tangent is passing through $(6,-5)$. Putting these values of x and y in equation of circle, we get,

$S_1={\left(6\right)}^2+{(-5)}^2-2\left(6\right)+4(-5)+3$

$\therefore S_1=36+25-12-20+3$

$\therefore S_1=32$

Now, by property of tangent, if T represents equation of pair of tangents, then we can write,

$S.S_1=T^2$

$\therefore T^2=32(x^2+y^2-2x+4y+3)$ Equation (1)

Now equation of tangent is given as,

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

$\therefore T:6x+(-5)y+(-1)(x+6)+2(y+(-5))+3=0$

$\therefore T:6x-y-x-6+2y-10+3=0$

$\therefore T:5x+y-13=0$

From equation (1), we can write,

${(5x+y-13)}^2=32(x^2+y^2-2x+4y+3)$

$\therefore 25x^2+y^2+169+10xy-26y-130x=32x^2+32y^2-64x+128y+96$

$\therefore -7x^2-31y^2+10xy-66x-154y+73=0$

Dividing equation by -1, we get,

$\therefore 7x^2+31y^2-10xy+66x+154y-73=0$

Answer is option (D)