Question

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

• A. $3(x^2+y^2)-14x-7y-14=0$
• B. $6(x^2+y^2)-14x-7y-14=0$
• C. $2(x^2+y^2)-14x-7y-14=0$
• D. None of these

Answer 1

The correct option is A $3(x^2+y^2)-14x-7y-14=0$

Let the equation of the circle be $P(x,y)=x^2+y^2+2gx+2fy+c=0$ ......$\left(1\right)$

The length of tangent from a point $(a,b)=\sqrt{P(a,b)}$

Accordingly the given tangents yield,

At $(1,0)$ is $1^2+0+2g+0+c=1$

$\Rightarrow 2g+c=0$ ......$\left(2\right)$

At $(0,2)$ is $0+2^2+0+4f+c={\left(\sqrt{7}\right)}^2$

$\Rightarrow 4f+c=7-4=3$ ......$\left(3\right)$

At $(2,2)$ is $3^2+2^2+6g+4f+c={\left(\sqrt{2}\right)}^2$

$\Rightarrow 6g+4f+c=2-9-4=-11$ ......$\left(4\right)$

Eliminating $g,f$ and $c$ we get

$\Rightarrow \left|\begin{array}{cccc}{x}^2+y^2& 2x& 2y& 1\\ 0& 2& 0& 1\\ -3& 0& 4& 1\\ 11& 6& 4& 1\end{array}\right|=0$

$\Rightarrow x^2+y^2\left|\begin{array}{ccc}2& 0& 1\\ 0& 4& 1\\ 6& 4& 1\end{array}\right|-2x\left|\begin{array}{ccc}0& 0& 1\\ -3& 4& 1\\ 11& 4& 1\end{array}\right|+2y\left|\begin{array}{ccc}0& 2& 1\\ -3& 0& 1\\ 11& 6& 1\end{array}\right|-1\left|\begin{array}{ccc}0& 2& 0\\ -3& 0& 4\\ 11& 6& 4\end{array}\right|=0$

$\Rightarrow (x^2+y^2)[2(4-4)-0+1(0-24)]-2x[0-0+1(-12-44)]$
$+2y[0-2(-3-11)+1(-18-0)]-1[0-2(-12-44)]=0$

$\Rightarrow -24(x^2+y^2)+112x+56y+112=0$

Divide both sides by $-8$ we get

$\Rightarrow 3(x^2+y^2)-14x-7y-14=0$