Question

Equation of the circle which is such that the lengths of the tangents to it from the points $(1,0),(2,0)$ and $(3,2)$ and $1,\sqrt{7},\sqrt{2}$ respectively is

• A. $6(x^2+y^2)-28x-5y+28=0$
• B. $9(x^2+y^2)-28x-5y+28=0$
• C. $3(x^2+y^2)-28x-5y+28=0$
• D. $x^2+y^2-28x-5y+28=0$

Answer 1

Answer: (A)

$6(x^2+y^2)-28x-5y+28=0$

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

Length of tangent $=\sqrt{5}$

Lengths given are $1,\sqrt{7},\sqrt{2}$

$\Rightarrow \sqrt{1+0+2g+0+c}=1$

$\Rightarrow \sqrt{2g+c+1}=1$

$2g+c+1=1$

$\Rightarrow c=-2g\to \left(1\right)$

$\sqrt{4+0+4g+c}=\sqrt{7}$

$\Rightarrow 4g+4+c=7$

$\Rightarrow 4g+c=3\to \left(2\right)$

$\sqrt{g+4+6g+4f+c}=\sqrt{2}$

$\Rightarrow 6g+4f+c+13=2$

$\Rightarrow 6g+4f+c=-11\to \left(3\right)$

Putting $\left(1\right)$ in $\left(2\right)$

$\Rightarrow 4g-2g=3$

$\Rightarrow g=\frac{3}{2}$

$c=-3$

Putting in $\left(3\right)$

$\Rightarrow 6\left(\frac{3}{2}\right)+4f-3=-11$

$\Rightarrow g+4f-3=-11$

$\Rightarrow 4f+6=-11$

$\Rightarrow f=\frac{-17}{4}$

Circle $x^2+y^2+3x-\frac{17}{2}y-3=0$