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

1 Answer
Barry H.
May 16, 2018

#y=4#

Explanation:

From the question, #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# , #d/dx[x^2+y^2=16]# = #2x+2ydy/dx=0#.

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

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

Hope this was helpful.

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