Question

lf the origin is shifted to the point $\left(\frac{ab}{a-b},0\right)$ without rotation then the equation
$(a-b)(x^2+y^2)-2abx=0$ becomes

• A. $(a-b)(x^2+y^2)-(a+b)xy+abx=a^2+b^2$
• B. $(a+b)(x^2+y^2)=2ab$
• C. $(a-b{)}^2(x^2+y^2)=a^2{b}^2$
• D. $(x^2+y^2)=(a^2+b^2)$

Answer 1

Answer: (C)

$(a-b{)}^2(x^2+y^2)=a^2{b}^2$

The origin is shifted to $(\frac{ab}{a-b},0)$

Replace $x$ $\to (x+\frac{ab}{a-b})$

$\Rightarrow (a-b)({(x+\frac{ab}{a-b})}^2+y^2)-2ab(x+\frac{ab}{a-b})=0$

$\Rightarrow (a-b)(x^2+\frac{a^2{b}^2}{(a-b{)}^2}+\frac{2abx}{(a-b)}+y^2)-2abx-\frac{2a^2{b}^2}{a-b}=0$

$\Rightarrow (a-b)x^2+(a-b)y^2+\left(\frac{a^2{b}^2-2a^2{b}^2}{a-b}\right)=0$

$\Rightarrow (a-b{)}^2(x^2+y^2)=a^2{b}^2$

Hence, option C is the correct answer.