Question

The lengths of the tangents from $(1,0),(2,0),(3,2)$ to a circle are $1,\sqrt{7},\sqrt{2}$. Then the equation of the circle is

• A. $x^2+y^2+6x+17y+6=0$
• B. $2x^2+2y^2+6x-17y-6=0$
• C. $2x^2+2y^2-6x-17y-6=0$
• D. $2x^2+2y^2-6x-17y+6=0$

Answer 1

Answer: (B)

$2x^2+2y^2+6x-17y-6=0$

Let equation of circle is $x^2+y^2+2gx+2fy+c=0$
$\therefore$ length of tangent from any point $P(x_1,y_1)$ to circle is
$P=\sqrt{S_1}=\sqrt{x_1^2+y_1^2+2g{x}_1+2f{y}_1+c}$
point $(1,0);L_1=1=\sqrt{1+2g+c}$
$2g+c=0...\left(1\right)$
from point $(2,0),L_2=\sqrt{7}=\sqrt{4+4g+c}$
$4g+c=3...\left(2\right)$
from point $(3,2),L_3=\sqrt{2}=\sqrt{13+6g+4f+c}$
$6g+4f+c+11=0...\left(3\right)$
From (1) & (2),
$g=\frac{3}{2},c=-3$
And from (3),
$9+4f-3+11=0$
$f=-\frac{17}{4}$
$\therefore$ equation of circle is
$x^2+y^2+2\times \frac{3}{2}x-2\times \frac{17}{4}y-3=0$
$\Rightarrow 2x^2+2y^2+6x-17y-6=0$