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Find the equations of those tangents to the circle $x^{2}+y^{2}=52,$ which are parallel to the straight line $2x+3y=6$.

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Toolbox:
  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
Let $(x_1,y_1)$ be the point at which the is $||$ to $ 2x+3y=6$
Therefore slope of the at $(x_1,y_1)$ is $m=\large\frac {-2}{3}$ (slope of the line)
Step 2:
$m=\large\frac{dy}{dx_{(x_1y_1)}}$
Differentiating the equation with respect to x
$2x+2y \large\frac{dy}{dx}$$=0$
Step 3:
$\large\frac{dy}{dx_{(x_1,y_1)}}=\large\frac{-x_1}{y_1}=\frac{-2}{3}$
=> $\large\frac{x_1}{\sqrt {52-{x_1}^2}}=\frac{2}{3}$
=>$3x_1=2 \sqrt{52-{x_1}^2}$
$9{x_1}^2=208-4{x_1}^2$
Therefore $13 {x_1}^2=208=>{x_1}^2=16=>x_1=\pm 6$
$y_1=\sqrt {52-x_1^2}=\sqrt {52-36}=\pm 4$
Step 4:
$x_1,y_1$ are of the same sign since $ \large\frac{x_1}{y_1} $$>0$
Therefore the points are (6,4),(-6,-4)
Step 5:
Equation of the tangent at (6,4)
$y-4=\large\frac{-2}{3}$$(x-6)$
$3y+12=-2x-12$
$2x+3y+24=0$
The tangent are $2x+3y\pm 24=0$
answered Jul 24, 2013 by meena.p
 

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