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The Power Rule can be proved using implicit diffe…

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Problem 47 Hard Difficulty

Show, using implicit differentiation, that any tangent line at a point $ P $ to a circle $ O $ is perpendicular to the radius $ OP. $

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Video Transcript

In this problem, we are asked to show that a radius of the circle so line o is perpendicular to tangent line. That has a straight point p. So, let's set at center of the circle is located directly 00, so that origin and point p is that x, not y, not all right. We know that the equation of the circle is x, squared plus y squared is r squared, and let's say that we are interested in the question of a tangent line to list perles differentiate this equation. With respect to x, we have 2 x plus 2 y times y prime is equal to 0. So from this we see that 2 will cancel out and a prime would then be equal to negative x over y. Now then, the equation of tangent line. We know that is often 4 y minus or not is equal to y prime x, not times x, minus x. Not we know what y prime is so, let's just put everything and we know that point p is a excelente 1. So we have y minus 1. Is equal to now the relative avolat given points would then be negative of x, not divided by y, not times x, minus x, not so. This is the equation of the tangent length. All right now, let's find an equation for a radius of line of line of p. Now that will be a form y minus 0 point. We are following this right here again, so y minus 0 is equal to. We need to find the slope, but we know the 2 points that line passes through so that will be y, not minus 0 divide x, not minus 0, that is the slope multiplied by x, minus 0 point. So if that is the slope of line of p and in this part, is the slope of the tangent line. If we were to multiply those 2 so negative x, not over y, not multiplied by y, not over x. Not we see that multiplication of the slopes gives us negative 1 point, so this means that those 2 lines are indeed perpendicular.

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