equation of tangent of circle∗
pahio†
2013-03-22 2:01:04
We derive the equation of tangent line for a circle with radius r. For simplicity, we chose for the origin the centre of the circle, when the points (x, y)
of the circle satisfy the equation
x2 + y 2 = r2 .
(1)
Let the point of tangency be (x0 , y0 ). Then the slope of radius with end
point (x0 , y0 ) is xy00 , whence, according to the parent entry, its opposite inverse
− xy00 is the slope of the tangent, being perpendicular to the radius. Thus the
equation of the tangent is written as
y − y0 = −
x0
(x − x0 ).
y0
Removing the denominator and the parentheses we obtain from this first x0 x +
y0 y = x20 + y02 , and then
x0 x + y0 y = r 2
(2)
since (x0 , y0 ) satisfies (1).
Remark. In the equation (2) of the tangent, x0 , y0 are the coordinates
of the point of tangency and x, y the coordinates of an arbitrary point of the
tangent line. But one can of course swap those meanings; then we interprete
(2) such that x0 , y0 are the coordinates of some fixed point P outside the circle
(1) and x, y the coordinates of the point of tangency of either of the tangents
which may be drawn from P to the circle. If (2) now is again interpreted as
an equation of a line (its degree is 1!), this line must pass through both the
mentioned points of tangency A and B (they satisfy the equation!); in a word,
(2) is now the equation of the tangent chord AB of P = (x0 , y0 ). See also
polar.
∗ hEquationOfTangentOfCirclei created: h2013-03-2i by: hpahioi version: h41254i Privacy
setting: h1i hDerivationi h51M20i h51M04i
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