Find the points on the curve $x^{2}+y^{2}=13$, the tangent at each one of which is parallel to the line $2 x+3 y=7$.
Let $\left(x_{1}, y_{1}\right)$ represent the required point.
The slope of line $2 x+3 y=7$ is $\frac{-2}{3}$.
Since, the point lies on the curve.
Hence, $x_{1}^{2}+y_{1}^{2}=13 \quad \cdots$ (1)
Now, $x^{2}+y^{2}=13$
On differentiating both sides w.r.t. $x$, we get
$2 x+2 y \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{-x}{y}$
Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-x_{1}}{y_{1}}$
Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the given line [Given]
$\Rightarrow \frac{-x_{1}}{y_{1}}=\frac{-2}{3}$
$\Rightarrow x_{1}=\frac{2 y_{1}}{3}$ .....(2)
From eq. (1), we get
$\left(\frac{2 y_{1}}{3}\right)^{2}+y_{1}^{2}=13$
$\Rightarrow \frac{13 y_{1}{ }^{2}}{9}=13$
$\Rightarrow y_{1}{ }^{2}=9$
$\Rightarrow y_{1}=\pm 3$
$\Rightarrow y_{1}=3$ or $y_{1}=-3$
and
$x_{1}=2$ or $x_{1}=-2$ [From eq. (2)]
Thus, the required points are $(2,3)$ and $(-2,-3)$