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tangent line

 Definition/Summary The tangent to a curve in a plane at a particular point has the same Gradient as the curve has at that point. More generally, the (n-1)-dimensional tangent hyperplane to an (n-1)-dimensional surface in n-dimensional space at a particular point has the same Gradient as the surface has at that point. So if $A\,=\,(a_1,a_2,\cdots a_n)$ is a point on a surface defined by the equation $F(x_1,x_2,\cdots x_n) = 0$, then the tangent hyperplane to the curve through $A$ is $\frac{\partial F}{\partial x_1}\arrowvert_A(x_1 - a_1)\,+\,\frac{\partial F}{\partial x_2}\arrowvert_A(x_2 - a_2)\,+\,\cdots\,\frac{\partial F}{\partial x_2}\arrowvert_A(x_n - a_n)\,=\,0$ If a curve in n dimensions is defined using a parameter t as $A(t)\,=\,(a_1(t),a_2(t),\cdots a_n(t))$ , then its tangent is: $(x_1 - a_1) / \frac{da_1}{dt}\,=\,(x_2 - a_2) / \frac{da_2}{dt}\,=\,\cdots\,=\,(x_n - a_n) / \frac{da_n}{dt}$

 Equations For example, if the point $A\,=\,(x_0,y_0)$ lies on the circle: $$F(x,y)\,=\,(x-p)^2+(y-q)^2\,-\,r^2\,=\,0$$ (1) then $$\frac{\partial F}{\partial x}\arrowvert_A\,=\,2(x-p)\arrowvert_A\,=\,2(x_0-p)$$ and $$\frac{\partial F}{\partial y}\arrowvert_A\,=\,2(y-q)\arrowvert_A\,=\,2(y_0-q)$$ and so the equation of the tangent at $A$ is: $$(x_0-p)(x-x_0)\,+\,(y_0-q)(y-y_0)=0$$ (2) Alternatively, the same circle can be defined by $A(\theta)\,=\,(p+r\cos\theta, q+r\sin\theta)$ and so the equation of the tangent at $A(\theta)$ is: $$\frac{x-p-r\cos\theta}{-r\sin\theta}\,=\,\frac{y-q-r\sin\theta}{r\cos\theta}$$

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 Breakdown Mathematics > Geometry >> Coordinate Geometry

 Extended explanation We will transform the equation (2) into more convenient type for better way of memorizing and using the formula. Because of $M(x_1,y_1) /in K$: $$(x_1-p)+(y_1-q)^2=r^2$$ (3) If we sum the equations (2) and (3), we get: $$(x_1-p)(x-x_1)+(y_1-q)(y-y_1)+(x_1-p)+(y_1-q)^2=r^2$$ $$(x_1-p)[x-x_1+x_1-p]+(y_1-q)[y-y_1+y_1-q]=r^2$$ $$(x_1-p)(x-p)+(y_1-q)(y-q)=r^2$$ (4) The equation (4) is equation of tangent of the circle in the point $M(x_1,y_1) \in K$. If the K have center (0,0), i.e $K: x^2+y^2=r^2$, then p=q=0, so the equation of the tangent is: $$x_1x+y_1y=r^2$$