Browse Questions

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

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 $2x+3y\pm 24=0$