How do you write an equation of the line tangent to the graph of the circle whose equation is $x^{2} + y^{2} = 16$ $x^2+y^2=16$ at the point (0,4)?

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How do you write an equation of the line tangent to the graph of the circle whose equation is $x^{2} + y^{2} = 16$ $x^2+y^2=16$ at the point (0,4)?

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Answer 1 · 2018-05-16T23:54:34

$y = 4$ $y=4$

Explanation:

From the question, $x^{2} + y^{2} = 16$ $x^2+y^2=16$ , we need to differentiate this expression to find the slope [gradient]

Differentiating both sides of the expression, implicitly w.r.t. $x$ $x$ , $\frac{d}{d x} [x^{2} + y^{2} = 16]$ $d/dx[x^2+y^2=16]$ = $2 x + 2 y \frac{d y}{d x} = 0$ $2x+2ydy/dx=0$ .

Rearranging, $\frac{d y}{d x} = - \frac{2 x}{2 y}$ $dy/dx = -[2x]/[2y$ = $- \frac{x}{y}$ $-x/y$ and so when $x = 0$ $x=0$ and $y = 4$ $y=4$ ,
$\frac{d y}{d x} = \frac{0}{4} = 0$ $dy/dx=0/4 = 0$ . [ the gradient is zero]
The equation of a line can be written as $[y - y 1] = m [x - x 1]$ $[y-y1]=m[x-x1]$ which gives,

$y - 4 = 0 [x - 0]$ $y-4=0[x-0]$ so, $y = 4$ $y=4$ is the equation of the tangent line to the circle $x^{2} + y^{2} = 16$ $x^2+y^2=16$ at the point $[0, 4]$ $[0,4]$

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How do you write an equation of the line tangent to the graph of the circle whose equation is $x^{2} + y^{2} = 16$ $x^2+y^2=16$ at the point (0,4)?

1 Answer

Explanation:

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Explanation:

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How do you write an equation of the line tangent to the graph of the circle whose equation is $x^{2} + y^{2} = 16$ $x^2+y^2=16$ at the point (0,4)?