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

Let the equation of the circle be $P(x, y)=x^{2}+y^{2}+2 g x+2 f y+c=0$

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

Accordingly the given tangents yield,

At $(1,0)$ is $1^{2}+0+2 \mathrm{~g}+0+\mathrm{c}=1$

$\Rightarrow 2 \mathrm{~g}+\mathrm{c}=0$

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

$\Rightarrow 4 \mathrm{f}+\mathrm{c}=7-4=3$

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

$\Rightarrow 6 \mathrm{~g}+4 \mathrm{f}+\mathrm{c}=2-9-4=-11$

From equation (2), $g=-\frac{c}{2}$ and from equation (3), $4 f=3-c$

Putting these values in equation (4), we get

$\Rightarrow 6 \times-\frac{c}{2}+3-c+c=-11 \Rightarrow-3 c=-14 \Rightarrow c=\frac{14}{3}$

So, $g=-\frac{7}{3}$ and $4 f=3-\frac{14}{3} \Rightarrow f=-\frac{5}{12}$

Hence, $P(x, y)=x^{2}+y^{2}-\frac{14}{3} x-\frac{5}{6} y+\frac{14}{3}=0$

$P(x, y)=6\left(x^{2}+y^{2}\right)-28 x-5 y+28=0$

Hence, option (A) is the correct answer.