Question

The point on a circle nearest to the point $P(2,1)$ is at a distance of $4units$ and the farthest point is $(6,5)$. Then find the equation of the circle.

Answer 1

Let x, y be the nearest & farthest point from $P(2,1)$

Given, $px=4$

$py=\sqrt{(6-2{)}^2+(5-1{)}^2}=4\sqrt{2}$

$xy=py-px=4\sqrt{2}-4$

$\Rightarrow \frac{xy}{xp}=\frac{4(\sqrt{2}-1)}{4}=\sqrt{2}-1=\left(\frac{m_1}{m_2}\right)$ (say)

Coordinates of $x=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

$=(\frac{6+2(\sqrt{2}-1)}{1+(\sqrt{2}-1)},\frac{5+(\sqrt{2}-1)}{1+(\sqrt{2}-1)})$

$=(\frac{4+2\sqrt{2}}{\sqrt{2}},\frac{\sqrt{2}+4}{\sqrt{2}})$

$=(2+2\sqrt{2},1+2\sqrt{2})$

Equation of circle$=(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$(x-6)(x-2-2\sqrt{2})+(y-5)(y-1-2\sqrt{2})=0$.