To prove $xx_{0} + yy_{0} = r ^ { 2 }$

Question

There is a circle ($x ^ {2} + y ^ {2} = r ^ { 2 }$), there is a point P ($x_{0}$, $y_{0}$) outside the circle, there are two lines tangent to the circle through the point P, and the tangent points are A($x_{1}$,$y_{1}$) and B($x_{2},y_{2}$) respectively. How do I prove that AB's linear equation is

$xx_{0} + yy_{0} = r ^ { 2 }$

(This question has been translated by machine)

My teacher gave me the solution is:

Because the point P is on two tangents

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So we have:

$x_{1}x_{0} + y_{1}y_{0} = r ^ { 2 }$

$x_{2}x_{0} + y_{2}y_{0} = r ^ { 2 }$

Then

$xx_{0} + yy_{0} = r ^ { 2 }$

I can't understand it.

Answer 1

We know that slope of tangent$=\frac{dy}{dx}$. Here differentiating both sides of the equation we get $\frac{dy}{dx}=\frac{-x}{y}$.So equation of tangent passing through the point $(x_1,y_1)$ is $$y-y_1=-\frac{x_1}{y_1}(x-x_1)$$ $$\Rightarrow xx_1+yy_1=x_{1}^2+y_{1}^2=r^2$$ Similarly for $(x_2,y_2)$ Now since $(x_0,y_0)$ lies on both the tangents you can get the two equations that your teachers gave by plugging the coordinates.

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Answer 2

AB is the chord of contact of the circle $x^2+y^2=r^2$ about the point P$(x_0,y_0)$.

Eq. of tangent to the circle at [pint A$(x_1,y_1)$ is $x_1x+y_1y=r^2$ as this passes through the point P, we have

$x_1x_0+y_1y_0=r^2....(1)$.

Next the Eq. of tangent to the same circle at point B$(x_2,y_2)$ is $x_2x+y_2y=r^2$ as this also passes through the point P, we have

$x_2 x_0+y_2 y_0=r^2.....(2)$

Eq. (1,2) are as though the equation of the line AB is $x_0x+y_0y=r^2$ and it passes through the points A and B, respectively.