Is the Tangent Line Equation Equal to the Derivative of the Circle Equation?

AI Thread Summary
The discussion revolves around proving that the line equation 2ax + 2by = a² + b² is tangent to the circle defined by 4x² + 4y² = a² + b². It emphasizes that differentiating equations is not the same as differentiating functions, highlighting the importance of using gradients to determine perpendicularity. The gradient of the circle's function points perpendicular to the circle, while the gradient of the line's function points perpendicular to the line itself. To find the tangent line, one can use implicit differentiation to derive dy/dx for the circle. The conversation suggests identifying the point where the line and circle intersect as a key step in confirming tangency.
Giuseppe
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Hello, I was wondering if anyone can help me with this problem.

Show that the line with equation 2ax+2by=a^2+b^2
is tangent to the circle with equation 4x^2+4y^2=a^2+b^2


If this is true, wouldn't the derivative of the circle equation be equal to the first equation? Would I just take the partial derivative with respect to X and then to Y?
 
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Be careful. You don't differentiate equations you differentiate functions. If you think of 4x2+ 4y2= a2+ b2 ans "f(x,y)= constant" then the partial derivatives of f form the grad f vector which points PERPENDICULAR to the circle, not tangent to it. Fortunately, if you do the same thing with the line (think of it as g(x,y)= a2+ b2 and find grad g) that will be perpendicular to the line so getting line is in the same direction as the tangent is just finding (x,y) so that those two vectors are in the same direction and both equations are satisfied.

Another way, perhaps simpler, is to find dy/dx for the circle by implicit differentiation and use that to find tangent lines.
 
have you spotted the point the line and circle coincide at? You should be getting that by inspection
 
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