What is the equation of the tangent to x^2 + y^2 - 8 = 0 at (2, 2).

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation of the tangent to the circle x^2 + y^2 - 8 = 0 at the point (2, 2) has to be determined.

The equation of the circle x^2 + y^2 - 8 = 0 can be written as (x - 0)^2 + (y - 2)^2 = 8

This gives the center of the circle as (0,0) and the radius is sqrt 8.

A line joining (0,0) to (2,2) is perpendicular to the tangent drawn at (2,2)

The slope of the line between (2,2) and (0,0) is y/x = 2/2 = 1

=> y = x

The slope of a line perpendicular to this is -1.

This gives the equation of the tangent as (y - 2)/(x - 2) = -1

=> y - 2 = 2 - x

=> x + y - 4 = 0

The equation of the tangent to the circle x^2 + y^2 - 8 = 0 at the point (2, 2) is x + y - 4 = 0

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