Question

A variable chord is drawn to the circle $ {{x}^{2}}+{{y}^{2}}-2ax=0 $ from the origin, then the locus of the centre of the circle which is made by taking the chord as the diameter, is

• A. $ {{x}^{2}}+{{y}^{2}}+ax=0 $
• B. $ {{x}^{2}}+{{y}^{2}}-ax=0 $
• C. $ {{x}^{2}}+{{y}^{2}}+ay=0 $
• D. $ {{x}^{2}}+{{y}^{2}}-ay=0 $

Answer 1

Answer: (B)

The given circle $ {{x}^{2}}+{{y}^{2}}-2ax=0 $ ...(i)
Let centre of the drawn circle be $ (\alpha ,\beta ) $ let variable chord $ y=mx $ ...(ii) On solving Eqs. (i) and (ii), we get
$ x=0,\frac{2a}{1+{{m}^{2}}} $ and $ y=0,\frac{2am}{1+{{m}^{2}}} $
$ \therefore $ Intersection points are
$ (0,0)\left( \frac{2a}{1+{{m}^{2}}},\frac{2am}{1+{{m}^{2}}} \right) $
Let coordinate of the centre of circle be $ (\alpha ,\beta ) $
$ \therefore $ $ \alpha =\frac{0+\frac{2a}{1+{{m}^{2}}}}{2}=\frac{a}{1+{{m}^{2}}}, $ and $ \beta =\frac{0+\frac{2am}{1+{{m}^{2}}}}{2}=\frac{am}{1+{{m}^{2}}}, $
Thus, $ \beta =\alpha m $
$ \Rightarrow $ $ m=\frac{\beta }{\alpha } $
Then, $ \alpha =\frac{a}{1+\frac{{{\beta }^{2}}}{{{\alpha }^{2}}}} $
$ \Rightarrow $ $ \alpha =\frac{a{{\alpha }^{2}}}{{{\alpha }^{2}}+{{\beta }^{2}}} $
$ \Rightarrow $ $ \alpha ({{\alpha }^{2}}+{{\beta }^{2}})=a{{\alpha }^{2}} $
$ \Rightarrow $ $ {{\alpha }^{2}}+{{\beta }^{2}}-a\alpha =0 $
Hence, locus of the centre of circle is $ {{x}^{2}}+{{y}^{2}}-ax=0 $