The point on the circle $x^{2}+y^{2}=2$ at the abscissa and ordinate increase at the same rate is
The point on the circle $x^{2}+y^{2}=2$ at the abscissa and ordinate increase at the same rate is
$(-1,-1)$
$(1,-1)$
$(1,1)$
$(-1,4)$
Solution
Given, equation of circle is
$x^{2}+y^{2}=2 \quad \text{...(i)}$
Taking derivative w.r.t. ' $t$ ' on both sides.
$2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}=0$
If abscissa and ordinate increase at the same rate, we have
$\frac{d x}{d t}=\frac{d y}{d t}$
$x \frac{d x}{d t}+y \frac{d x}{d t}=0 \Rightarrow \frac{d x}{d t}(x+y)=0$
Since,
$\Rightarrow \quad x+y=0 \Rightarrow x=-y \quad \text{...(ii)}$
Solving Eqs. (i) and (ii), we get
$x^{2}+(-x)^{2}=2 \Rightarrow x=\pm 1$
For $x=1, y=-1$ and $x=-1, y=1$
Required point are $(1,-1)$ and $(-1,1)$.