# Find at what points on the circle $x^{2}+y^{2}=13$, the tangent is parallel to the line $2x+3y=7$.

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) Step 1: Let (x_1,y_1) be the point at which the tangent is || to 2x+3y=7 Therefore slope of the tangent = slope of the line =\large\frac{-2}{3}$$(=m)$
Step 2:
$m=\large\frac{dy}{dx_{(x_1,y_1)}}$
Differentiating $x^2+y^2=13$ with respect to x
$2x+2y \large\frac{dy}{dx}$$=0=>\large\frac{dy}{dx}=\frac{-x}{7}$
$m=\large\frac{-x_1}{y_1}=\frac{2}{3}$
$\large\frac{x_1}{\sqrt {13-x_1}^2}=\frac{-2}{3}$
$3x_1=-2 \sqrt {13-{x_1}^2}$
=>$9{x_1}^2=52-4{x_1}^2$
=>$13{x_1}^2=52\; or\; {x_1}^2=4$
Therefore $x_1=\pm 2$
$y_1=\sqrt {13-{x_1}^2}=\sqrt {13-4}=\pm 3$
$x_1,y_1$ are of opposite signs since $m <0$ and $m=\large\frac{-x_1}{y_1}$
Step 3:
Therefore the 2 points are $(2,-3)$ and$(-2,3)$