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

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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$(-2,3)$

 

answered Jul 24, 2013 by meena.p
 

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