verify for all points (x1,y1) on circle x^2+y^2=r^2, that all points (x2,y2) on tangent line to circle at (x1,y1) satisfy x1x2+y1y2=r^2

Question

assume (x₁,y₁) is a point on the circle x²+y²=r²

the tangent line will be -x₁/y₁=(y-y₁)/(x-x₁)

-x(x₁)+x₁²=y(y₁)-y₁²

x(x₁)+y(y₁)=r²

this means all points (x₂,y₂) on the tangent line for point (x₁,y₁) on circle radius r, center at origin

will always satisfy equation

x₁x₂+y₁y₂=r²

To find point on tangent line for point (x₁,y₁) on circle x²+y²=r²

choose any x₂, then y₂=(r²-x₁x₂)/y₁

Answer 1

Okay, let's verify the derivation step-by-step.

1. The Circle: We start with the equation of a circle centered at the origin (0,0) with radius r: x2+y2=r2
2. A Point on the Circle: Let (x1​,y1​) be any point on this circle. This means that the coordinates (x1​,y1​) must satisfy the circle's equation: x12​+y12​=r2 (Equation 1)
3. Slope of the Radius: The radius connects the center (0,0) to the point (x1​,y1​). The slope of this radius, mradius​, is: mradius​=x1​−0y1​−0​=x1​y1​​ (assuming x1​=0)
4. Slope of the Tangent Line: The tangent line to the circle at point (x1​,y1​) is perpendicular to the radius at that point. The slope of the tangent line, mtangent​, is the negative reciprocal of the slope of the radius: mtangent​=−mradius​1​=−y1​x1​​ (assuming y1​=0)
5. Equation of the Tangent Line: We can use the point-slope form of a line, y−y0​=m(x−x0​), with the point (x1​,y1​) and the slope mtangent​=−y1​x1​​. Let (x,y) be any point on the tangent line. y−y1​=−y1​x1​​(x−x1​)
6. Simplifying the Equation: Multiply both sides by y1​ (assuming y1​=0): y1​(y−y1​)=−x1​(x−x1​) Distribute: y1​y−y12​=−x1​x+x12​ Rearrange the terms to bring the x and y terms to one side: x1​x+y1​y=x12​+y12​
7. Using the Circle Equation: From Equation 1, we know that x12​+y12​=r2. Substitute this into the equation we just derived: x1​x+y1​y=r2
8. Verification: This equation, x1​x+y1​y=r2, represents the tangent line to the circle x2+y2=r2 at the point (x1​,y1​). Any point (x2​,y2​) that lies on this tangent line must satisfy this equation. Therefore, if we replace the general point (x,y) on the line with a specific point (x2​,y2​) on the line, the equation must hold: x1​x2​+y1​y2​=r2

Handling Special Cases:

1. Case 1: y1​=0. If y1​=0, then from x12​+y12​=r2, we get x12​=r2, so x1​=±r. The point is either (r,0) or (−r,0).
2. At (r,0), the tangent line is vertical, with the equation x=r. Any point (x2​,y2​) on this line has x2​=r. Let's check the formula x1​x2​+y1​y2​=r2: (r)(r)+(0)y2​=r2⟹r2=r2. This holds.
3. At (−r,0), the tangent line is vertical, with the equation x=−r. Any point (x2​,y2​) on this line has x2​=−r. Let's check the formula x1​x2​+y1​y2​=r2: (−r)(−r)+(0)y2​=r2⟹r2=r2. This also holds.
4. Case 2: x1​=0. If x1​=0, then from x12​+y12​=r2, we get y12​=r2, so y1​=±r. The point is either (0,r) or (0,−r).
5. At (0,r), the tangent line is horizontal, with the equation y=r. Any point (x2​,y2​) on this line has y2​=r. Let's check the formula x1​x2​+y1​y2​=r2: (0)x2​+(r)(r)=r2⟹r2=r2. This holds.
6. At (0,−r), the tangent line is horizontal, with the equation y=−r. Any point (x2​,y2​) on this line has y2​=−r. Let's check the formula x1​x2​+y1​y2​=r2: (0)x2​+(−r)(−r)=r2⟹r2=r2. This also holds.

Conclusion:

Your derivation and conclusion are correct. For any point (x1​,y1​) on the circle x2+y2=r2, the equation of the tangent line at that point is given by x1​x+y1​y=r2. Consequently, any point (x2​,y2​) lying on this tangent line must satisfy the equation: x1​x2​+y1​y2​=r2

Your final statement about finding y2​ given x2​ by rearranging the formula to y2​=(r2−x1​x2​)/y1​ is also correct, provided y1​=0. If y1​=0, the tangent line is vertical (x2​=x1​=±r), and x2​ cannot be chosen arbitrarily.

Answer 2

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