Question

If the length of the tangent from $(f,g)$ to the circle $x^2+y^2=6$ be twice the length of the tangent from the same points to the circle $x^2+y^2+3x+3y=0$, then

• A. $4f+4g+1=0$
• B. $4f-4g+1=0$
• C. $\left(f^2\right)+g^2+4f+4g+2=0$
• D. $\left(f^2\right)+g^2-4f-4g+2=0$

Answer 1

Answer: (C)

$\left(f^2\right)+g^2+4f+4g+2=0$

We know that, the length of the tangent from point

$P(x_1,y_1)$ to the circle $x^2+y^2+2gx+2fy+c=0$

is equal to

$\sqrt{x_1^2+y_1^2+2g{x}_1+2f{y}_1+c}$

So,

length of tangent from $(f,g)$ to the circle $x^2+y^2=6$

$=\sqrt{f^2+g^2-6}$

also,

length of tangent from $(f,g)$ to the circle

$x^2+y^2+3x+3y=0$

$=\sqrt{f^2+g^2+3f+3g}$

Now, according to question,

$\sqrt{f^2+g^2-6}=2\sqrt{f^2+g^2+3f+3g}$

squaring an both sides,

$f^2+g^2-6=2^2(f^2+g^2+3f+3g)$

$f^2+g^2-6=4(f^2+g^2+3f+3g)$

$f^2+g^2-6=4f^2+4g^2+12f+12g$

$4f^2-f^2+4g^2-g^2+12f+12g+6=0$

$3f^2+3g^2+12f+12g+6=0$

$3(f^2+g^2+4y+4g+2)=0$

$f^2+g^2+4f+4g+2=0$