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A circle of radius 3 is centered at the origin and passes through the point .(2,sqrt5) (a) Find an equation for the line through the origin and the point . (b) Find an equation for the tangent line to the circle at . (2,sqrt5)
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1) Find the slope line going through \((0, 0)\) and \((2,\sqrt{5})\) 2) The slope of a perpendicular line is just the negative reciprocal.
once you have the slopes of each you can use point slope formula
if it is centered at the origin and has radius 3, then the equation must be \(x^2+y^2=9\)
if you can use calculus, this is easy if not you do what @wio said
actually @wio answer is easiest nvm
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