STOP CIRCLING AROUND

Geometry Level 2

You have a circle which is based on the function x 2 + y 2 = 1 x^{2} + y^{2} = 1 . A line is tangent to that circle on the point, ( 3 2 , 1 2 ) \frac{\sqrt{3}}{2},\frac{1}{2}) . What is the y-intercept of the line tangent to the circle ?


The answer is 2.

Anthony Ling by Anthony Ling , 24, USA

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2 solutions

Differentiate the given equation of circle wrt dx to find slope of tangent at the given point then, just use point slope form of line to find the equation of tangent then, put x=0 to find y-intercept

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It might be helpful to mention the term 'implicit differentiation' here. Rather than solving the original equation one can do this, as you have suggested.

d d x ( x 2 + y 2 − 1 ) = 2 x + 2 y d y d x \frac { d }{ dx } \left( { x }^{ 2 }+{ y }^{ 2 }-1 \right) =2x+2y\frac { dy }{ dx }

Then solve this for the derivative of y wrt x.

d y d x = − x y \frac { dy }{ dx } =-\frac { x }{ y }

We have been offered the value (x,y), hence we have the slope of the tangent.

Bill Bell - 6 years, 10 months ago
Suraj Lal
Aug 16, 2014

equation of line = > y = mx + c

m = d y / d x ( x 2 + y 2 − 1 ) = − x / y h e r e x = 3 / 2 & y = 1 / 2 m = − 3 y = m x + c c = 1 / 2 − [ ( − 3 ) × 3 / 2 ] c = 2 y = − 3 x + 2 t a k e x = 0 f o r g e t t i n g y i n t e r c e p t y = 2 m\quad =\quad dy/dx({ x }^{ 2 }+{ y }^{ 2 }-1)\quad =\quad -x/y\\ here\quad x=\sqrt { 3 } /2\quad \& \quad y=\quad 1/2\\ m\quad =\quad -\sqrt { 3 } \\ y=\quad mx\quad +\quad c\\ c\quad =\quad 1/2\quad -\left[ \left( -\sqrt { 3 } \right) \quad \times \quad \sqrt { 3 } /2 \right] \\ c\quad =\quad 2\\ y\quad =\quad -\sqrt { 3 } x\quad +\quad 2\\ take\quad x\quad =\quad 0\quad for\quad getting\quad y\quad intercept\\ y\quad =\quad 2

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