Question

A circle passes through $(0,0)$ and $(1,0)$ and touches the circle $x^2+y^2=9$ , then the centre of circle is:

• A. $[\frac{3}{2},\frac{1}{2}]$
• B. $[\frac{1}{2},\frac{3}{2}]$
• C. $[\frac{1}{2},\frac{1}{2}]$
• D. $[\frac{1}{2},\pm \sqrt{2}]$

Answer 1

The correct option is D $[\frac{1}{2},\pm \sqrt{2}]$

Let Circle be:

$x^2+y^2+2gx+2fy+c=0(\because (U,0\left)liesoncircle\right)\Rightarrow 0+0+0+0+c=0\Rightarrow c=0.\because (1,0)liesonittoo\Rightarrow 1+0+2g+0+0=0\Rightarrow g=\frac{-1}{2}.Centreofcircleis(-g,-f)\Rightarrow (\frac{1}{2},-f)radius=\sqrt{g^2+f^2-c}=\sqrt{\frac{1}{4}+f^2}\because Ittouchescircle{x}^2+y^2=9Ifcircletouchesexternally:\therefore Distancebetween(0,0)and(\frac{1}{2},-f)=\sqrt{\frac{1}{4}+f^2}+3.\Rightarrow Butherecircletouchesinternally.\Rightarrow Distancebetween(0,0)and(\frac{1}{2},-f)=3-\sqrt{\frac{1}{4}+f^2}\Rightarrow \sqrt{\frac{1}{4}+f^2}=3-\sqrt{\frac{1}{4}+f^2}\Rightarrow 2\sqrt{\frac{1}{4}+f^2}=3\Rightarrow \frac{1}{4}+f^2=\frac{9}{4}.\Rightarrow f^2=\frac{8}{4}=2.\therefore f=\pm \sqrt{2}.Hence,centreis\to (-g,-f)\Rightarrow (\frac{1}{2},\pm \sqrt{2})\therefore D)isthecorrectoption.$