The equation of a circle is #(x- 3)^2 + (y +2)^2= 25#. The point (8 -2) is on the circle What is the equation of the line that is tangent to the circle at (8,-2)?

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Answer 1 · 2016-12-05T00:46:50

Here is an answer that uses no calculus.

The center of the circle is #(3,-2)#.

The line segment joining #(3,-2)# and #(8,-2)# is a radius. The slope of this line is #m=0# (It is a horizontal line.)

The tangent is perpendicular to the radius,

so the tangent is a vertical line. It goes through the point #(8,-2)# so its equation is #x=8#.

Answer 2 · 2016-12-05T01:24:24

For contrast, here is an answer that does use calculus.

Start by expanding the equation of the circle.

#x^2 - 6x + 9 + y^2 + 4y + 4 = 25#

Use the power rule and implicit differentiation to find the derivative.

#2x - 6 + 2y(dy/dx) +4(dy/dx) = 0#

#2y(dy/dx) + 4(dy/dx) = 6- 2x#

#dy/dx(2y + 4) = 6 - 2x#

#dy/dx= (2(3 - x))/(2(y + 2)#

#dy/dx= -(x - 3)/(y + 2)#

The slope of the curve at #(x, y)# is given by evaluating #(x, y)# in the derivative.

#m_"tangent" = -(8 + 3)/(-2 + 2)#

#m_"tangent" = -11/0#

The tangent line will therefore be vertical, or of the form #x = a#, where #a# is the x-coordinate in the given point #(x, y)#.

The equation is therefore #x= 8#.

Hopefully this helps!